Question

How do you find the center and radius of the circle ${{x}^{2}}+{{y}^{2}}-4x+12y-36=0$?

Answer 1

The general form of the equation of circle is ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$. We can find the coordinates of the center of the circle and the radius using the values of the coefficient of the equation. The coordinates of the centre of the circle are $\left( -g,-f \right)$. The radius of the circle is $\sqrt{{{g}^{2}}+{{f}^{2}}-c}$. To find the values of the coefficient, we need to compare the given equation with the general form of the equation of circle.

Complete step by step solution:
We are given the equation of the circle ${{x}^{2}}+{{y}^{2}}-4x+12y-36=0$. We need to find its center and radius. Comparing the given equation with the general form of the equation of the circle. We get the values of g, f, and c as $-2,6\And -36$ respectively. We know that for a circle represented by the general form of the equation, the X and Y coordinates of the center are $-g,-f$ respectively. Thus, for the given circle, by substituting the values of g and f, we get the coordinates of the center as $\left( -(-2),-6 \right)$. Simplifying this, we get centre coordinates $\left( 2,-6 \right)$.
The radius of circle is $\sqrt{{{g}^{2}}+{{f}^{2}}-c}$, substituting the values of g, f and c, we get radius as $\sqrt{{{2}^{2}}+{{\left( -6 \right)}^{2}}-\left( -36 \right)}=2$.
We can plot the circle as,

Note: We can also use the centre radius form of the equation which is expressed as ${{\left( x-a \right)}^{2}}+{{\left( y-b \right)}^{2}}={{r}^{2}}$. Here, a and b are the X-coordinate and Y-coordinate of the center, and r is the radius of the circle. But this form will require simplifying the equation which may get difficult.