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# High School Geometry - Equation of a Circle

## Introduction

• The locus of a point that moves in a plane such that its distance from a fixed point is always constant is called a Circle.
• The fixed point is called the center, and the constant distance is called the circle's radius.
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## Equation of a Circle

Central form of a circle:

• The equation of a circle with its center at C(h, k) and radius r units is given by:

${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={r}^{2}$

Simplest form of a circle:

• If the center of a circle is at the origin, the equation of the circle reduces to its simplest form:

${x}^{2}+{y}^{2}={r}^{2}$

## Solved Examples

Question 1: Find the equation of a circle with its center at $\left(-3,-8\right)$ and having a radius of 9 units.

Solution: ${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={r}^{2}$

Here, $h=-3$$k=-8$ and $r=9$ units.

Equation of the circle:

${\left[x-\left(-3\right)\right]}^{2}+{\left[y-\left(-8\right)\right]}^{2}={\left(9\right)}^{2}$

${\left(x+3\right)}^{2}+{\left(y+8\right)}^{2}=81$

Question 2: Identify the coordinates of the center and radius of a circle represented by the equation ${x}^{2}+{y}^{2}-4x-10y+13=0$.

Solution: ${x}^{2}+{y}^{2}-4x-10y+13=0$

${x}^{2}-4x+{y}^{2}-10y+13=0$

$\left({x}^{2}-4x+4\right)-4+\left({y}^{2}-10y+25\right)-25+13=0$

${\left(x-2\right)}^{2}+{\left(y-5\right)}^{2}-16=0$

So, the center of the given circle is at (2, 5) and its radius is 4 units.

Question 3: Find the equation of a circle whose center is $\left(3,-2\right)$ and which passes through the point .

Solution: The equation of the circle with the center $\left(3,-2\right)$ and radius $r$ is:

${\left(x-3\right)}^{2}+{\left(y+2\right)}^{2}={r}^{2}$

${\left(6-3\right)}^{2}+{\left(2+2\right)}^{2}={r}^{2}$

${\left(3\right)}^{2}+{\left(4\right)}^{2}={r}^{2}$

${\left(x-3\right)}^{2}+{\left(y+2\right)}^{2}={\left(5\right)}^{2}$

${x}^{2}-6x+9+{y}^{2}+4y+4=25$

${x}^{2}+{y}^{2}-6x+4y=12$

Question 4: Find the equation of a circle whose diameters are $2x-y=6$ and $x+2y=8$ and the area is .

Solution: The point of intersection of the diameters $2x-y=6$ and $x+2y=8$ is . Thus the center of the circle will be at .

The radius of the circle can be found using the given area of the circle.

${\mathrm{\pi r}}^{2}\mathrm{=}154$

$\frac{22}{7}{\mathrm{r}}^{2}=154$

${\mathrm{r}}^{2}=\frac{154×7}{22}=49$

Therefore, the equation of the circle will be: ${\left(x-4\right)}^{2}+{\left(y-2\right)}^{2}=49$.

## Cheat Sheet

• The equation of a circle with center  and radius $\mathbit{r}$ units is:

${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={r}^{2}$

## Blunder Area

• Attention must be paid to the sign of numbers while substituting the values in the equation of the circle.