Question

The centre of the circle, whose radius is $5$ and which touches the circle $x^2+y^2-2x-4y-20=0$ at $\left(5,5\right)$ is

• A. $\left(10,5\right)$
• B. $\left(5,8\right)$
• C. $\left(5,10\right)$
• D. $\left(8,9\right)$
• E. $\left(9,8\right)$

Answer 1

The correct option is E

$\left(9,8\right)$

Explanation for the correct option:

Finding the center:

Given equation of circle is

$$\begin{gathered} x^2+y^2-2x-4y-20=0 \\ \Rightarrow x^2-2x+1+y^2-4y+4-20=5 \\ \Rightarrow {\left(x-1\right)}^2+{\left(y-2\right)}^2=25 \\ \Rightarrow {\left(x-1\right)}^2+{\left(y-2\right)}^2=5^2 \end{gathered}$$

Therefore, the center of this circle is $\left(1,2\right)$.

Now since the radius of each circle is $5$, the point at which both circles touch each other divides the line segment joining both centers into exactly half.

Let $\left(h,k\right)$ be the required center of circle.

By midpoint formula,

$$\begin{gathered} \left(5,5\right)=\left(\frac{1+h}{2},\frac{k+2}{2}\right) \\ \Rightarrow \left(h,k\right)=\left(9,8\right) \end{gathered}$$

Therefore, the correct answer is option (D).