One of my math tests has this question.
A circle has its center at $(6,7)$ and goes through $(1,4)$. Another circle is tangent at $(1,4)$ and has the same area.
What are the possible coordinates of the second circle. Show your work or explain how you found you answer.
Could someone solve a sample question to show me how to solve this one?
The center $Q=(h,k)$ of the second circle lies on the line defined by $P_1=(6,7)$ and $P_2=(1,4)$. What means
$$\frac{k-4}{h-1}=\frac{4-7}{1-6}=\frac{3}{5}...(1)$$
Since these circles have the same area it follows they have equal radius. Then $$\sqrt{(h-1)^2+(k-4)^2}=\sqrt{(1-6)^2+(4-7)^2}...(2)$$
Solving the system of equations formed by $(1)$ and $(2)$ you find the center of the second circle.