Question

The equation of a common tangent to be circle $x^2+y^2=16$ and to the Ellipse $\frac{x^2}{49}+\frac{y^2}{4}=1$ is

• A. $y=x+\sqrt{45}$
• B. $y=x+\sqrt{53}$
• C. $\sqrt{11}y=2x+4$
• D. $\sqrt{11}y=2x+4\sqrt{15}$

Answer 1

Answer: (D)

$\sqrt{11}y=2x+4\sqrt{15}$

$x^2+y^2=16$

Equation of tangent to circle in slope form is

$y=mx\pm \sqrt[r]{r1+m^2}$

$y=mx\pm \sqrt[4]{41+m^2}$ (1)

+Equation of tangent to euipse$\frac{x^2}{49}+\frac{y^2}{4}\Rightarrow$

Is $y=mx\pm \sqrt{a^2{m}^2+b^2}$

$y=mx\pm \sqrt{49m^2+4}$ (2)

(1) + (2) must be same so

$4\sqrt{1+m^2}=\sqrt{49m^2+4}$

$\Rightarrow 16\left(1+m^2\right)=\pm \left(49m^2+4\right)$

$\Rightarrow 16+16m^2=49m^2+4$

$\Rightarrow 33m^2=12\Rightarrow m=\pm \sqrt{\frac{4}{11}}=\frac{2}{\sqrt{11}}$

$y=\sqrt{\frac{4}{11}x}+4\sqrt{1+\frac{4}{11}}$

$=\sqrt{\frac{4}{11}x}+\frac{4\sqrt{15}}{\sqrt{11}}$

$\sqrt{11}y=2x+4\sqrt{15}$