Question

How do you find the center and radius of the circle \[4{{x}^{2}}-16x+4{{y}^{2}}-24y+36=0\]?

Answer 1

Hint: The general form of the equation of circle is \[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\]. We can see that coefficients of the square terms are one in the above equation. In the given equation, this condition is not followed, so we have to make the coefficients one. We can find the coordinates of the center of the circle and the radius using the values of the coefficient of the equation. The coordinate of the center 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 \[4{{x}^{2}}-16x+4{{y}^{2}}-24y+36=0\].
Dividing both sides of the equation by 4, we get \[{{x}^{2}}-4x+{{y}^{2}}-6y+9=0\]. 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 \[-2,-3\And 9\] 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 \[\left( -(-2),-(-3) \right)\]. Simplifying this, we get center coordinates \[\left( 2,3 \right)\].
The radius of the circle is \[\sqrt{{{g}^{2}}+{{f}^{2}}-c}\], substituting the values of g, f, and c, we get radius as\[\sqrt{{{2}^{2}}+{{3}^{2}}-9}=2\].
The graph of the circle is as follows,

Note: We should remember for an equation of conic of form \[a{{x}^{2}}+b{{y}^{2}}+2gx+2fy+2hxy+c=0\] it must follow the following conditions to represent a circle. The first condition is \[a=b\]. The second condition is that \[h\] must be zero.