Question

The radius of the circle passing through $A \left(\frac{15}{4}, \sqrt{7}\right)$ and $B (3,0)$ which touches the auxiliary circle of the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$, is

• A. $\frac{\sqrt{121}}{4}$
• B. $\frac{\sqrt{121}}{8}$
• C. $\sqrt{51}$
• D. $\frac{\sqrt{337}}{4}$

Answer 1

Answer: (B)

The point $A\left(\frac{15}{4}, \sqrt{7}\right)$ lies on the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ and $B(3,0)$ is one of the foci of the ellipse.
Circle with $AB$ as diameter touches the auxiliary circle of the ellipse.
$\therefore \text { Radius of the circle }=\frac{1}{2} \sqrt{\left(\frac{15}{4}-3\right)^2+7}=\frac{1}{2} \times \frac{\sqrt{121}}{4}=\frac{\sqrt{121}}{8} $