Question

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

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

Answer 1

Answer: (E)

$(9,8)$

Let the two circles be $C_1$ and $C_2$

Equation of circle $C_1$ is $x^2+y^2-2x-4y-20=0\text{or}(x-1{)}^2+(y-2{)}^2=5^2$.

Hence, for $C_1$ coordinates of the centre and radius are $(1,2)$ and $5$ respectively

We know that radius of $C_2$ is also $5$. Hence the point of contact of the 2 circles, $(5,5)$, divides the line segment joining the centres in the ratio $1:1$.

Hence, if the coordinate of centre of $C_2$ are $(x_1,y_1)$, then $\frac{x_1+1}{2}=5⟹x_1=9\text{and}\frac{y_1+2}{2}=5⟹y_2=8$

Hence, coordinates of centre of circle $C_2$ is $(9,8)$