Question

The common tangent to the circles $x^2+y^2=4$ and $x^2+y^2+6x+8y-24=0$ intersects the coordinate axes at $A$ and $B$ respectively. If $OA$ and $OB$ are equal to half of the length of the major and minor axes of an ellipse respectively, where $O$ is the origin, then the eccentricity of the ellipse is

• A. $\frac{\sqrt{5}}{4}$
• B. $\frac{\sqrt{3}}{4}$
• C. $\frac{\sqrt{7}}{4}$
• D. $\frac{1}{\sqrt{2}}$

Answer 1

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

Given circles are circles $x^2+y^2=4$ and $x^2+y^2+6x+8y-24=0$
Center and radius of the circles are
$C_1=(0,0),r_1=2C_2=(-3,-4),r_2=\sqrt{9+16+24}=7$
Therefore,
$C_1{C}_2=\sqrt{9+16}=5,|r_1-r_2|=5$
As $C_1{C}_2=|r_1-r_2|$, so the circles touches each other internally.

The equation of common tangent is
$S_1-S_2=0\Rightarrow 6x+8y-20=0\Rightarrow 3x+4y-10=0\Rightarrow A\equiv (\frac{10}{3},0),B\equiv (0,\frac{10}{4})$
$OA=\frac{10}{3}=\frac{1}{2}\times 2aOB=\frac{10}{4}=\frac{1}{2}\times 2b\Rightarrow a=\frac{10}{3},b=\frac{10}{4}\Rightarrow e^2=1-\frac{b^2}{a^2}\therefore e=\frac{\sqrt{7}}{4}$