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: $x^2+y^2-4x-10y+13=0$

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

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

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

${\left(x-2\right)}^2 + {\left(y-5\right)}^2={\left(4\right)}^2$

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 $\left(6, 2\right)$.

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

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

$\text{Circle passes through }\left(6, 2\right)$

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

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

$r^2=9+16 =25$

$r=\sqrt{25}=5 \text{units}$

$\text{Therefore, the equation of circle is:}$

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

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

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

 

Question 4: Find the equation of a circle whose diameters are $2x-y=6$ and $x+2y=8$ and the area is $154 {\text{unit}}^2$.

Solution: The point of intersection of the diameters $2x-y=6$ and $x+2y=8$ is $\left(4, 2\right)$. Thus the center of the circle will be at $C\left(4, 2\right)$.

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

$\text{Area}=154 {\text{unit}}^2$

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

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

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

$\mathrm{r}=\sqrt{49}=7 \text{units}$

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 $\left(h, k\right)$ and radius $\mathit{r}$ 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.