Question

Find the equation of the common tangent in $1^{st}$ quadrant to the circle $x^2 + y^2 = 16$ and the ellipse $\frac{x^2}{25} + \frac{y^2}{4} = 1$. Also find the length of the intercept of the tangent between the coordinate axes.

Answer 1

The equation of the common tangent is $2x + \sqrt{3}y - 4\sqrt{21} = 0$, and the length of the intercept is 14.

Answer 2

The equation of the circle is $x^2 + y^2 = 16$ (radius $r=4$). The equation of the ellipse is $\frac{x^2}{25} + \frac{y^2}{4} = 1$ ($a=5, b=2$).

The general equation of a tangent to the circle is $y = mx \pm 4\sqrt{1+m^2}$. The general equation of a tangent to the ellipse is $y = mx \pm \sqrt{25m^2+4}$.

For a common tangent, $16(1+m^2) = 25m^2+4$. Solving for $m^2$: $16 + 16m^2 = 25m^2 + 4 \implies 12 = 9m^2 \implies m^2 = \frac{4}{3} \implies m = \pm \frac{2}{\sqrt{3}}$.

For the tangent in the 1st quadrant, we need $m < 0$ and a positive y-intercept. Let $m = -\frac{2}{\sqrt{3}}$. The y-intercept $c$ satisfies $c^2 = 25m^2+4 = 25(\frac{4}{3})+4 = \frac{100}{3} + \frac{12}{3} = \frac{112}{3}$. So, $c = \sqrt{\frac{112}{3}} = \frac{4\sqrt{7}}{\sqrt{3}} = \frac{4\sqrt{21}}{3}$ (taking the positive root for the y-intercept).

The equation of the tangent is $y = -\frac{2}{\sqrt{3}}x + \frac{4\sqrt{21}}{3}$, which simplifies to $2x + \sqrt{3}y - 4\sqrt{21} = 0$.

The x-intercept (set $y=0$) is $2x = 4\sqrt{21} \implies x_0 = 2\sqrt{21}$. The y-intercept (set $x=0$) is $\sqrt{3}y = 4\sqrt{21} \implies y_0 = \frac{4\sqrt{21}}{\sqrt{3}} = 4\sqrt{7}$.

The length of the intercept $L$ is $\sqrt{x_0^2 + y_0^2} = \sqrt{(2\sqrt{21})^2 + (4\sqrt{7})^2} = \sqrt{84 + 112} = \sqrt{196} = 14$.