Equation of a Circle Calculator (General and Standard Form, Center, Radius, Circle and Line)

Choose the direction of the conversion and enter the coefficients of the circle. The equation below is linked to the input fields, so you can also edit the coefficients in the equation directly. To check how a line meets the circle, choose "Also check the line px + qy + s = 0" in the last drop-down.

Conversion

x coefficient l

y coefficient m

Constant n

Center x-coordinate a

Center y-coordinate b

Right side (radius squared r²)

Circle and a line

Line's x coefficient p

Line's y coefficient q

Line's constant s

Enter numbers only. Decimals, negative numbers and fractions such as 3/4 are OK. A blank coefficient of x or y counts as 1 (x is the same as 1x), and a blank constant or center coordinate counts as 0. Enter 0 for terms you do not use.

Result and graph

Enter the coefficients in the fields on the left and press "Calculate". The result and a figure of the circle will appear here.

[ CLOSE ]

What you can do on this page
What is this calculation used for?
How to Use
Formulas and figures
Symbols and terms
- Symbols
- Terms
Good to know before you start
How to calculate it in Excel
How to calculate it in Google Sheets
How to calculate it in Python
How to write it in LaTeX and other math languages (copy and paste)
How to have ChatGPT do the calculation
DataChef Features
Related Features
NumberChef Calculators List

What you can do on this page

Complete the square to turn the general form of a circle \(x^{2} + y^{2} + lx + my + n = 0\) into standard form \((x - a)^{2} + (y - b)^{2} = r^{2}\), and get the center and radius as exact values with fractions and \(\sqrt{\ }\) (with steps)
You can also go the other way: expand from the center and the right side (the radius squared) to get the general form
If the right side \(r^{2}\) is 0 or less, the equation is not a circle (it is a single point, or there is no graph). The calculator tells you which, and why
It also tells you how a line \(px + qy + s = 0\) meets the circle (at two points, tangent, or not at all) by comparing the distance \(d\) from the center to the line with the radius \(r\), and gives the intersection points as exact values
The calculator shows the equation of the circle the way you would write it by hand, and you can edit the coefficients right in the equation. You can check the result on a figure of the circle and the line

Coordinates and coefficients can be decimals, negative numbers or fractions such as 3/4.

What is this calculation used for?

Collision detection in games (circular hitboxes)

In 2D games, it is standard to model the hitbox of a character or a bullet as a circle. Whether a bullet's path (a line) hits an enemy's circle is decided exactly by this page's check: compare the distance \(d\) from the center to the line with the radius \(r\). Whether two circles overlap is decided the same way, by comparing the distance between their centers with the sum of their radii.
Circles are used because their shape does not change when they turn, and one distance calculation settles the check. If you learn game programming, you will see equations of circles at work in every frame.

Modeling signal coverage (cell towers and Wi-Fi)

Under ideal conditions with no obstacles, the area a cell tower or Wi-Fi router can reach is modeled as a circle centered at the antenna (radius = the range). "Is this spot in range?" is a comparison of the distance from the center with the radius, in other words, a check of whether the spot is inside or outside the circle.
Real signals are affected by terrain and buildings, so they do not follow this exactly. Still, circle models are a starting point for deciding where to put towers so that there are no gaps in coverage.

The idea behind positioning (GPS and your phone's location)

GPS positioning is based on this idea: once you know your distance from a satellite, the places you could be narrow down to a sphere (a circle, if you think in a plane). Solving the circles (spheres) of distances from several satellites together gives the point where they meet, which is your location.
The real calculation is in 3D and also corrects for clock errors, so it is more complicated. But the idea of the equation of a circle, "the set of points at a fixed distance", is its foundation.

Design, drafting and CAD (managing arcs and hole positions)

Drawings of machine parts and buildings are full of circles: bolt holes, curved edges, cross sections of pipes and so on. CAD software stores a circle as "the coordinates of the center and the radius", which is the information in the standard form, and it calculates where circles and lines cross, and tangent lines, every time you edit a shape.
Checks like "if I drill a hole here, is it far enough from the edge?" also come down to calculating distances from a center.

Designing curves on roads and railways (circular curves)

The basic shape of a curve on a road or railway is an arc of a circle, called a circular curve, and design drawings manage it by its center and radius. The smaller the radius, the sharper the curve, so road design standards set a minimum curve radius for each design speed.
Where a straight section joins a curve smoothly, the condition "the line is tangent to the circle" is used directly. The idea of a tangent line shapes real roads.

Formulas and figures

Equation of a circle (standard form)

Figure

Standard notation (the usual math form)

\((x - a)^{2}\) \(+\) \((y - b)^{2}\) \(=\) \(r^{2}\)

In words (symbols replaced with words)

① square of the difference between \(x\) and the center's \(a\) \(+\) ② square of the difference between \(y\) and the center's \(b\) \(=\) ③ \(r^{2}\): radius squared

The formula in words

① For any point \((x,\ y)\) on the circle, the square of the \(x\) distance from the center, \(x - a\)

② plus the square of the \(y\) distance from the center, \(y - b\)

③ equals the square of the radius \(r\)

Quick example

The equation of the circle with center (1, 2) and radius 3 is

\((x - 1)^{2}\) \(+\) \((y - 2)^{2}\) \(=\) \(3^{2} = 9\)

\((x - 1)^{2} + (y - 2)^{2} = 9\)

Key idea

A circle is the set of all points at a fixed distance (the radius \(r\)) from the center. Write the distance between a point \((x,\ y)\) on the circle and the center \((a,\ b)\) with the Pythagorean theorem, and you get \(\sqrt{(x-a)^{2} + (y-b)^{2}} = r\). Squaring both sides gives the standard form. In other words, the standard form is the definition of a circle, "distance \(r\) from the center", written as an equation. Note that the right side is \(r^{2}\), the radius squared. The radius of \((x-1)^{2} + (y-2)^{2} = 9\) is not 9 but \(\sqrt{9} = 3\).

General form → standard form (finding the center and radius by completing the square)

Standard notation (the usual math form)

\(\mathrm{C}\) \(=\) \(\left(-\dfrac{l}{2},\ -\dfrac{m}{2}\right)\)

\(r^{2}\) \(=\) \(\dfrac{l^{2}}{4} + \dfrac{m^{2}}{4} - n\)

In words (symbols replaced with words)

② \(\mathrm{C}\): center \(=\) ① half of the \(x\) coefficient \(l\) and of the \(y\) coefficient \(m\), with signs changed

④ \(r^{2}\): radius squared \(=\) ③ sum of the squares of the half coefficients, minus the constant \(n\)

The formula in words

① Complete the square on the general form \(x^{2} + y^{2} + lx + my + n = 0\). The halves of the coefficients \(l\) and \(m\), with signs changed,

② give the center \(\mathrm{C}\)

③ and the sum of the squares of the half coefficients minus the constant \(n\)

④ is the radius squared \(r^{2}\)

Quick example

Completing the square on \(x^{2} + y^{2} - 4x - 6y + 9 = 0\) gives

center \((2,\ 3)\) \(,\) \(r^{2} = 4 + 9 - 9 = 4\)

\(x^{2} - 4x = (x - 2)^{2} - 4\)

\(y^{2} - 6y = (y - 3)^{2} - 9\)

\((x - 2)^{2} + (y - 3)^{2} = 4 + 9 - 9 = 4\)

\(r = \sqrt{4} = 2\)

Key idea

Completing the square rewrites "a squared term plus a linear term", such as \(x^{2} - 4x\), as "(something) squared plus a constant", such as \((x - 2)^{2} - 4\). You build \((x - 2)^{2}\) from half the \(x\) coefficient (half of \(-4\) is \(-2\)). Expanding it gives an extra \(+4\), so you subtract that amount to keep the two sides equal. The right side of the result is \(r^{2}\). Always check its sign here. If the right side is 0, only one point, the center, satisfies the equation. If it is negative, no point satisfies it at all, because a sum of squares is never negative. So an equation in general form does not always describe a circle. It is a circle only when the right side is positive.

How a line meets a circle (comparing the distance d from the center to the line with the radius r)

Figure

Standard notation (the usual math form)

\(d\) \(=\) \(\dfrac{\left|\,pa + qb + s\,\right|}{\sqrt{p^{2} + q^{2}}}\)

In words (symbols replaced with words)

② \(d\): distance from the center \((a,\ b)\) to the line \(=\) ① absolute value of the line's left side at the center, divided by \(\sqrt{p^{2} + q^{2}}\)

The formula in words

① For the line \(px + qy + s = 0\), the absolute value of the left side with the center \((a,\ b)\) plugged in, divided by \(\sqrt{p^{2} + q^{2}}\),

② is the distance \(d\) from the center to the line . Compare this \(d\) with the radius \(r\) to see how the line meets the circle

Quick example

For the circle \((x - 2)^{2} + (y - 3)^{2} = 4\) (center \((2,\ 3)\), radius 2) and the line \(3x + 4y - 8 = 0\)

distance \(d = 2\) \(=\) radius \(r = 2\)

\(d = \dfrac{|3 \times 2 + 4 \times 3 - 8|}{\sqrt{3^{2} + 4^{2}}} = \dfrac{10}{5} = 2\)

\(d = r\)

Key idea

There are three cases, depending on which is larger, the distance \(d\) or the radius \(r\). If \(d < r\), the line crosses the circle at two different points. If \(d = r\), it touches the circle at exactly one point (the line is a tangent line, and the shared point is the point of tangency). If \(d > r\), they do not meet. Comparing numbers with \(\sqrt{\ }\) is easy to get wrong, so this calculator compares \(d^{2}\) and \(r^{2}\) as exact fractions. You can use the same trick by hand. Another method is to solve the equations of the circle and the line together: eliminate \(y\) (or \(x\)) and look at the sign of the discriminant \(D\) of the resulting quadratic equation (\(D > 0\): two points, \(D = 0\): tangent, \(D < 0\): no points). To find the coordinates of the intersection points, you solve this system.

The equation of a circle can be written in two forms: the standard form \((x - a)^{2} + (y - b)^{2} = r^{2}\), which shows the center and radius, and the expanded general form \(x^{2} + y^{2} + lx + my + n = 0\). Completing the square turns the general form into the standard form, but it is a circle only when the right side (the value in the place of \(r^{2}\)) is positive. How a line meets a circle is decided by comparing the distance \(d\) from the center to the line with the radius \(r\).

Symbols and terms

Symbols

\((x,\ y)\)	x, y	The coordinates of a point moving on the circle. The equation of a circle says: "collect every \((x,\ y)\) that makes this true, and you get the circle".
\(a,\ b\)	a, b	The letters for the center \((a,\ b)\) of the circle. Letters from the start of the alphabet are usually used for fixed numbers. Unlike the moving \(x\) and \(y\), they stay fixed for one circle. Many US textbooks write the center as \((h,\ k)\).
\(r\)	are	The radius of the circle, from the first letter of "radius". Note that the right side of the standard form is not \(r\) itself but its square \(r^{2}\).
\(\mathrm{C}\)	see	A letter often used for the center of a circle, from the first letter of "center". The circle itself is sometimes named \(C\) too.
\(l,\ m,\ n\)	l, m, n	The coefficients of the general form \(x^{2} + y^{2} + lx + my + n = 0\). Many textbooks write them as \(D,\ E,\ F\) (\(x^{2} + y^{2} + Dx + Ey + F = 0\)); the meaning is the same.
\(d\)	dee	The distance from the center to the line, from the first letter of "distance". Comparing it with the radius \(r\) tells you how the line meets the circle.
\(p,\ q,\ s\)	p, q, s	The coefficients of the line \(px + qy + s = 0\) on this page. A line is usually written \(ax + by + c = 0\), but this page uses other letters so they do not clash with the center \((a,\ b)\).
\(\sqrt{\ }\)	square root (radical sign)	The sign for the positive number whose square is the number inside (the square root). The radius is found as \(r = \sqrt{r^{2}}\). For example, if \(r^{2} = 8\), then \(r = \sqrt{8} = 2\sqrt{2}\).

Terms

circle	The figure made of all the points in a plane at a fixed distance (the radius) from one point (the center). The equation of a circle is this "fixed distance" definition written as an equation.
equation of a circle	An equation that describes a circle on the coordinate plane. The points \((x,\ y)\) that satisfy it make up exactly the circle. It is taught in high school Geometry and Algebra 2.
standard form	The form \((x - a)^{2} + (y - b)^{2} = r^{2}\) (often written \((x - h)^{2} + (y - k)^{2} = r^{2}\)). Its strength is that you can read the center \((a,\ b)\) and the radius \(r\) just by looking at it.
general form	The form \(x^{2} + y^{2} + lx + my + n = 0\), made by expanding and simplifying the standard form. You cannot read the center or radius directly, but since it is already expanded, it is easy to work with during calculations.
completing the square	Rewriting an expression such as \(x^{2} + lx\) as "(something) squared plus a constant", as in \(\left(x + \dfrac{l}{2}\right)^{2} - \dfrac{l^{2}}{4}\). It is also used to find the vertex of a quadratic function, and it is the core technique of this page.
center	The middle point of a circle, the same distance (the radius) from every point on the circle. The \((a,\ b)\) in the standard form is the center.
radius	The distance \(r\) from the center to a point on the circle. As a length, it is always positive. It is found by taking the square root of the right side \(r^{2}\) of the standard form.
Pythagorean theorem	The theorem that in a right triangle, the square of the hypotenuse equals the sum of the squares of the other two sides. It is the basis of the distance between two points, and the standard form of a circle is this theorem written as an equation. It is taught in Grade 8.
intersection point	A point that lies on both of two figures. For a circle and a line, there are 2 when the line crosses the circle, 1 when it is tangent, and 0 when it misses. The coordinates are found by solving the two equations together.
tangent	A line is tangent to a circle when they share exactly one point. The line is called a tangent line, and the shared point is the point of tangency. The distance from the center to a tangent line is exactly the radius.
chord	When a line crosses a circle at two points, the segment between those points is a chord. The perpendicular from the center to a chord cuts it exactly in half, so half the chord, the distance \(d\) from the center to the line and the radius \(r\) are linked by the Pythagorean theorem.
distance from a point to a line	The length of the perpendicular from a point to a line (the shortest distance from the point to the line). It is calculated with \(d = \dfrac{\|ax_0 + by_0 + c\|}{\sqrt{a^{2} + b^{2}}}\) and is used to decide how a line meets a circle.
discriminant	The expression \(D = b^{2} - 4ac\) that tells how many solutions the quadratic equation \(ax^{2} + bx + c = 0\) has. The discriminant of the quadratic you get by solving the circle and line equations together also tells how they meet (\(D > 0\): two points, \(D = 0\): tangent, \(D < 0\): no points).
coordinate plane	A plane made by crossing a horizontal number line (the \(x\)-axis) and a vertical number line (the \(y\)-axis) at a right angle, so that every point can be written as a pair of numbers \((x,\ y)\). It is where geometry and equations meet.

Good to know before you start

Here is what helps you use the calculation on this page with real understanding, not just by pressing the button.
If you get stuck, going back to review these topics is the fastest way forward.

The coordinate plane and coordinates of points (Grades 5–6)	Being able to write the position of a point as a pair of numbers \((x,\ y)\) (for example, the point \((2,\ 3)\) is 2 right and 3 up from the origin) Having a feel for the idea that "all the points that satisfy an equation make up its graph (a figure)"
The Pythagorean theorem and the distance formula (Grade 8)	Knowing that in a right triangle, the square of the hypotenuse equals the sum of the squares of the other two sides Knowing that the distance between two points is \(\sqrt{(x_2 - x_1)^{2} + (y_2 - y_1)^{2}}\) (this is the Pythagorean theorem itself)
Square roots (Grade 8 and Algebra 1)	Being able to find the positive number whose square is a given number, as in \(\sqrt{4} = 2\) Being able to simplify a root by taking out square factors, as in \(\sqrt{8} = 2\sqrt{2}\)
Expanding, factoring and completing the square (Algebra 1 and Algebra 2)	Being able to expand expressions such as \((x - 2)^{2} = x^{2} - 4x + 4\) Being able to complete the square, as in \(x^{2} - 4x = (x - 2)^{2} - 4\) (the same step used to find the vertex of a quadratic function) Being able to solve quadratic equations by factoring or the quadratic formula, and knowing how the sign of the discriminant \(D = b^{2} - 4ac\) relates to the number of solutions
Distance from a point to a line (Geometry and Precalculus)	Being able to write a line as \(ax + by + c = 0\) (general form), including rewriting it from \(y = mx + k\) Knowing that the distance from the point \((x_0,\ y_0)\) to the line \(ax + by + c = 0\) is \(\dfrac{\|ax_0 + by_0 + c\|}{\sqrt{a^{2} + b^{2}}}\)

How to calculate it in Excel

Copy the whole table below and paste it into cell A1 in Excel. It works as is.

Table for general form → standard form (center and radius)

x coefficient l	-4
y coefficient m	-6
Constant n	9
Center x-coordinate (−l/2)	=-B1/2
Center y-coordinate (−m/2)	=-B2/2
Radius squared r² (l²/4 + m²/4 − n)	=B1^2/4+B2^2/4-B3
Radius r (a circle only if r² > 0)	=IF(B6>0,SQRT(B6),"not a circle")

Table for standard form → general form

Center x-coordinate a	1
Center y-coordinate b	-2
Right side (radius squared r²)	9
x coefficient l (−2a)	=-2*B1
y coefficient m (−2b)	=-2*B2
Constant n (a² + b² − r²)	=B1^2+B2^2-B3

Table to check how a line meets the circle

Center x-coordinate a	2
Center y-coordinate b	3
Radius r	2
Line's x coefficient p	3
Line's y coefficient q	4
Line's constant s	-8
Distance d from center to line	=ABS(B4B1+B5B2+B6)/SQRT(B4^2+B5^2)
Result	=IF(B7<B3,"two points",IF(B7=B3,"tangent","no points"))

After pasting, the upper rows (coefficients and so on) are your inputs and the lower rows are calculated automatically. "^" is a power, SQRT gives the square root, and ABS gives the absolute value.
The first table is the example x²+y²−4x−6y+9=0: center (2, 3), r²=4 and radius 2. If r² is 0 or less, the equation is not a circle, and the cell shows "not a circle".
The second table is the example with center (1, −2) and r²=9: l=−2, m=4, n=−4, so the equation is x²+y²−2x+4y−4=0.
The third table is the example of the circle (x−2)²+(y−3)²=4 and the line 3x+4y−8=0. d=2 equals the radius 2, so the result is "tangent". SQRT can have small decimal errors, so for an exact tangent check it is safer to compare d² and r² (compare =(B4*B1+B5*B2+B6)^2/(B4^2+B5^2) with B3^2).

How to calculate it in Google Sheets

Copy the whole table below and paste it into cell A1 in Google Sheets. It works as is.

Table for general form → standard form (center and radius)

x coefficient l	-4
y coefficient m	-6
Constant n	9
Center x-coordinate (−l/2)	=-B1/2
Center y-coordinate (−m/2)	=-B2/2
Radius squared r² (l²/4 + m²/4 − n)	=B1^2/4+B2^2/4-B3
Radius r (a circle only if r² > 0)	=IF(B6>0,SQRT(B6),"not a circle")

Table for standard form → general form

Center x-coordinate a	1
Center y-coordinate b	-2
Right side (radius squared r²)	9
x coefficient l (−2a)	=-2*B1
y coefficient m (−2b)	=-2*B2
Constant n (a² + b² − r²)	=B1^2+B2^2-B3

Table to check how a line meets the circle

Center x-coordinate a	2
Center y-coordinate b	3
Radius r	2
Line's x coefficient p	3
Line's y coefficient q	4
Line's constant s	-8
Distance d from center to line	=ABS(B4B1+B5B2+B6)/SQRT(B4^2+B5^2)
Result	=IF(B7<B3,"two points",IF(B7=B3,"tangent","no points"))

The same formulas as in Excel work as is. Copy the whole table, paste it into cell A1, and replace the coefficients with your own numbers.

How to calculate it in Python

from fractions import Fraction
import math

# Coefficients of the general form x² + y² + lx + my + n = 0 (a fraction such as 3/4 can be written as Fraction(3, 4))
coef_l = Fraction(-4)
coef_m = Fraction(-6)
coef_n = Fraction(9)

center_x = -coef_l / 2
center_y = -coef_m / 2
radius_squared = coef_l ** 2 / 4 + coef_m ** 2 / 4 - coef_n

print(f"Center: ({center_x}, {center_y})")
print(f"Radius squared r²: {radius_squared}")
if radius_squared > 0:
    print(f"Radius r: {math.sqrt(radius_squared)}")
elif radius_squared == 0:
    print("Not a circle: it describes only one point, the center")
else:
    print("Not a circle: no point satisfies this equation")

# Position relative to the line px + qy + s = 0 (d² and r² are compared as fractions, so there is no rounding error)
line_p = Fraction(3)
line_q = Fraction(4)
line_s = Fraction(-8)
distance_squared = (line_p * center_x + line_q * center_y + line_s) ** 2 / (line_p ** 2 + line_q ** 2)
if distance_squared < radius_squared:
    print("The circle and the line meet at two different points")
elif distance_squared == radius_squared:
    print("The line is tangent to the circle")
else:
    print("The circle and the line do not meet")

With the fractions module from the standard library, the calculation stays in exact fractions with no rounding error. This example uses the circle x²+y²−4x−6y+9=0 and the line 3x+4y−8=0. Running it shows the center (2, 3), r²=4, the radius 2.0 and "The line is tangent to the circle". The distance and the radius are compared as d² and r² so that decimal errors from the square root cannot throw off the tangent check. Change the coefficients and run it again.

How to write it in LaTeX and other math languages (copy and paste)

Equation of a circle (standard form)

(x − a)² + (y − b)² = r²

(x - a)^{2} + (y - b)^{2} = r^{2}

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <msup><mrow><mo>(</mo><mi>x</mi><mo>&#x2212;</mo><mi>a</mi><mo>)</mo></mrow><mn>2</mn></msup>
    <mo>+</mo>
    <msup><mrow><mo>(</mo><mi>y</mi><mo>&#x2212;</mo><mi>b</mi><mo>)</mo></mrow><mn>2</mn></msup>
    <mo>=</mo>
    <msup><mi>r</mi><mn>2</mn></msup>
  </mrow>
</math>

(x - a)^2 + (y - b)^2 = r^2

(x - a)^2 + (y - b)^2 == r^2

(x - a)^2 + (y - b)^2 = r^2;

(x - a)^2 + (y - b)^2 == r^2

(x - a)^2 + (y - b)^2 = r^2

General form → standard form (finding the center and radius by completing the square)

(x + l/2)² + (y + m/2)² = l²/4 + m²/4 − n

\left(x + \frac{l}{2}\right)^{2} + \left(y + \frac{m}{2}\right)^{2} = \frac{l^{2}}{4} + \frac{m^{2}}{4} - n

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <msup><mrow><mo>(</mo><mi>x</mi><mo>+</mo><mfrac><mi>l</mi><mn>2</mn></mfrac><mo>)</mo></mrow><mn>2</mn></msup>
    <mo>+</mo>
    <msup><mrow><mo>(</mo><mi>y</mi><mo>+</mo><mfrac><mi>m</mi><mn>2</mn></mfrac><mo>)</mo></mrow><mn>2</mn></msup>
    <mo>=</mo>
    <mfrac><msup><mi>l</mi><mn>2</mn></msup><mn>4</mn></mfrac>
    <mo>+</mo>
    <mfrac><msup><mi>m</mi><mn>2</mn></msup><mn>4</mn></mfrac>
    <mo>&#x2212;</mo>
    <mi>n</mi>
  </mrow>
</math>

(x + l/2)^2 + (y + m/2)^2 = l^2/4 + m^2/4 - n

Simplify[(x + l/2)^2 + (y + m/2)^2 - (l^2/4 + m^2/4 - n) == x^2 + y^2 + l x + m y + n]

Student:-Precalculus:-CompleteSquare(x^2 + y^2 + l*x + m*y + n, [x, y]);

expand((x + l/2)^2 + (y + m/2)^2 - (l^2/4 + m^2/4 - n))

(x + l/2)^2 + (y + m/2)^2 = l^2/4 + m^2/4 - n

How a line meets a circle (comparing the distance d from the center to the line with the radius r)

d = |pa + qb + s| / √(p² + q²)

d = \frac{\left|pa + qb + s\right|}{\sqrt{p^{2} + q^{2}}}

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <mi>d</mi>
    <mo>=</mo>
    <mfrac>
      <mrow><mo>|</mo><mi>p</mi><mi>a</mi><mo>+</mo><mi>q</mi><mi>b</mi><mo>+</mo><mi>s</mi><mo>|</mo></mrow>
      <msqrt><mrow><msup><mi>p</mi><mn>2</mn></msup><mo>+</mo><msup><mi>q</mi><mn>2</mn></msup></mrow></msqrt>
    </mfrac>
  </mrow>
</math>

d = |p a + q b + s| / sqrt(p^2 + q^2)

d = Abs[p a + q b + s]/Sqrt[p^2 + q^2]

d := abs(p*a + q*b + s)/sqrt(p^2 + q^2);

d = abs(p*a + q*b + s)/sqrt(p^2 + q^2);

d = |pa + qb + s|/√(p^2 + q^2)

How to have ChatGPT do the calculation

You are a math assistant for coordinate geometry. Do the following calculation by actually running Python code, and base your answer only on the numbers from the execution result (do not answer by mental math or guessing).

For the circle x² + y² − 4x − 6y + 9 = 0, show each of the following:
1. The equation rewritten in standard form (x−a)² + (y−b)² = r² by completing the square
2. The center and the radius (as exact values)
3. How the line 3x + 4y − 8 = 0 meets the circle (at two different points / tangent / not at all), and the coordinates of any intersection points

In Python, use the fractions module from the standard library (or sympy) to calculate exactly, and compare the distance with the radius by comparing d² with r². Show the formulas you used and the numbers from the execution result.

How to Use

1

Enter your numbers

Type the numbers you want to calculate with into the input fields
2

Calculate

Press the "Calculate" button
3

Check the result

The result appears on the spot. The same page also explains the idea behind the calculation and the formula