Question

the point $(-3, 6)$ is on a circle with a center at $(2, 3)$. what is the equation of the circle?
\bigcirc $(x - 2)^2 + (y - 3)^2 = 16$
\bigcirc $(x + 2)^2 + (y + 3)^2 = 34$
\bigcirc $(x - 2)^2 + (y - 3)^2 = 34$
\bigcirc $(x + 2)^2 + (y + 3)^2 = 16$

Explanation:

Step1: Recall the circle equation

The standard equation of a circle with center \((h,k)\) and radius \(r\) is \((x - h)^2+(y - k)^2=r^2\). Here, the center is \((2,3)\), so \(h = 2\) and \(k = 3\). The equation becomes \((x - 2)^2+(y - 3)^2=r^2\).

Step2: Calculate the radius

The point \((-3,6)\) lies on the circle, so we can substitute \(x=-3\) and \(y = 6\) into the equation to find \(r^2\).
Substitute into \((x - 2)^2+(y - 3)^2=r^2\):
\((-3 - 2)^2+(6 - 3)^2=r^2\)
\((-5)^2+(3)^2=r^2\)
\(25 + 9=r^2\)
\(34=r^2\)

Step3: Write the final equation

Substitute \(r^2 = 34\) and \((h,k)=(2,3)\) into the circle equation: \((x - 2)^2+(y - 3)^2=34\).

Answer:

\(\boldsymbol{(x - 2)^2+(y - 3)^2=34}\) (corresponding to the option: \((x - 2)^2+(y - 3)^2=34\))