Question

The point at which the circle $x^2+y^2-4x-4y+7=0$ and $x^2+y^2-12x-10y+45=0$ touch each other is

• A. $(\frac{14}{5},\frac{13}{5})$
• B. $(\frac{2}{5},\frac{5}{6})$
• C. $(\frac{14}{5},\frac{13}{6})$
• D. $(\frac{12}{5},2+\frac{\sqrt{21}}{5})$

Answer 1

The correct option is A $(\frac{14}{5},\frac{13}{5})$

The center of the first circle is $(2,2)$ and its radius is $\sqrt{4+4-7}=1$

The center of the second circle is $(6,5)$ and its radius is $\sqrt{36+25-45}=4$

Distance between centers = $\sqrt{(6-2{)}^2+(5-2{)}^2}=5=1+4$

Thus, the line joining $(2,2)$ and $(6,5)$ is divided internally in the ratio $1:4$ by the intersection point.

The point is thus $(\frac{8+6}{1+4},\frac{8+5}{1+4})=(\frac{14}{5},\frac{13}{5})$