If y=m_1x+c_1 and y=m_2x+c_2, m_1 m_2, are two common tangents of the circle x^2+y^2=2 and…

Question

If $y=m_1x+c_1$ and $y=m_2x+c_2$, $m_1
e m_2$, are two common tangents of the circle $x^2+y^2=2$ and the parabola $y^2=x$, then the value of $8|m_1m_2|$ is equal to:

Accepted Answer

Tangent to $y^2=x$: $y=mx+\dfrac1{4m}$. Tangent to the circle:
$$\frac{1/(4m)}{\sqrt{1+m^2}}=\sqrt2\ \Longrightarrow\ 32m^4+32m^2-1=0\ \Longrightarrow\ m^2=\frac{-4+3\sqrt2}8$$
(taking the positive root). By symmetry the two common tangents have slopes $\pm m$, so $|m_1m_2|=m^2$:
$$8|m_1m_2|=8m^2=-4+3\sqrt2.$$

Solutions · 1