Question

The length of a common tangent to $x^2+y^2=16$ and $9x^2+25y^2=225$ is

• A. $\frac{9}{4}$
• B. $\frac{\sqrt{3}}{4}$
• C. $\frac{3}{4}\sqrt{7}$
• D. $\frac{5}{4}\sqrt{7}$

Answer 1

The correct option is C $\frac{3}{4}\sqrt{7}$

Ellipse is $\frac{x^2}{25}+\frac{y^2}{9}=1$

Any point on the ellipse is $(5\cos \theta ,3\sin \theta )$ at which the tangent is $\frac{x}{5}\cos \theta +\frac{y}{3}\sin \theta =1$
If it is a tangent to circle $x^2+y^2=16$, then $p=r$ gives $\frac{1}{\sqrt{(\frac{{\cos }^2\theta }{25}+\frac{{\sin }^2\theta }{9})}}=4$ $\therefore \frac{{\cos }^2\theta }{25}+\frac{1-{\cos }^2\theta }{9}=\frac{1}{16}\therefore \cos \theta =\frac{5\sqrt{7}}{16}$ $l^2=25{\cos }^2\theta +9{\sin }^2\theta -16=16{\cos }^2\theta -7=16\cdot \frac{25.7}{{\left(16\right)}^2}-7=\frac{175-112}{16}=\frac{63}{16}\therefore l=\frac{3}{4}\sqrt{7}\Rightarrow \left(c\right).$