Question

B) there is a line with positive slope that is tangent to both circles. find the equation of this line, and (c) determine the points at which this tangent line touches each circle

Answer 1

Given that there is a line with positive slope that is tangent to both circles $x^2 + y^2 = 1$ and $(x-3)^2 + y^2 = 4$.

Let the point of tangency in the first circle be (a, b) and the point of tangency in the second circle be (c, d), then the slope of the tangent to the first circle is given by

$2x+\left.2y \frac{dy}{dx}\right|_{(a, b)} =0 \\ \\ \Rightarrow \left.\frac{dy}{dx}\right|_{(a, b)} =- \frac{x}{y} =- \frac{a}{b}$

and the slope of the second circle is given by

$2(x-3)+\left.2y \frac{dy}{dx}\right|_{(c, d)} =0 \\ \\ \Rightarrow \left.\frac{dy}{dx}\right|_{(a, b)} =- \frac{x-3}{y} =- \frac{c-3}{d}$

From the first circle equation, we have

$x^2 = 1 - y^2 \\ \\ \Rightarrow a^2 = 1 - b^2$

and from the second equation, we have

$(x-3)^2 = 4 - y^2 \\ \\ (c-3)^2 = 4 - d^2$

Since the two slopes above are for the same line, then

$\frac{a}{b}= \frac{c-3}{d} \\ \\ \Rightarrow ad=b(c-3) \\ \\ \Rightarrow a^2d^2=b^2(c-3)^2 \\ \\ \Rightarrow(1-b^2)d^2=b^2(4-d^2) \\ \\ \Rightarrow d^2-b^2d^2=4b^2-b^2d^2 \\ \\ \Rightarrow d^2=4b^2 \\ \\ \Rightarrow d=\pm2b$

Thus, we have

$\frac{a}{b}= \frac{c-3}{2b} \\ \\ \Rightarrow a= \frac{c-3}{2}$