Circles
Externally Tangent Circles
nta_pyq_2024_apr
Grade 11
Question:
Let the circles $C_1:(x-\alpha)^2+(y-\beta)^2=r_1^2$ and $C_2:(x-8)^2+\left(y-\dfrac{15}{2}\right)^2=r_2^2$ touch each other externally at the point $(6,6)$. If the point $(6,6)$ divides the line segment joining the centres of the circles $C_1$ and $C_2$ internally in the ratio $2:1$, then $(\alpha+\beta)+4(r_1^2+r_2^2)$ equals:
Step-by-Step Solution
Key Concept: $(6,6)$ divides $C_1(\alpha,\beta)$ and $C_2(8,15/2)$ in ratio $2:1$: $\frac{16+\alpha}{3}=6$ and $\frac{15+\beta}{3}=6\Rightarrow(\alpha,\beta)=(2,3)$.
$(\alpha,\beta)=(2,3)$, $r_1=5$, $r_2=5/2$. $(2+3)+4(25+25/4)=5+4\cdot125/4=5+125=130$.
Correct Answer: 2