Let the equations of two diameters of a circle x^2+y^2-2x+2fy+1=0 be 2px-y=1 and 2x+py=4p.

Question

Let the equations of two diameters of a circle $x^2+y^2-2x+2fy+1=0$ be $2px-y=1$ and $2x+py=4p$. Then the slope $m\in(0,\infty)$ of the tangent to the hyperbola $3x^2-y^2=3$ passing through the centre of the circle is equal to:

Answer: 2

Accepted Answer

Centre $(1,-f)$. Both diameters pass through it:
$$2p+f=1,\qquad 2-pf=4p.$$

Solving: $2p^2-5p+2=0\Rightarrow p=2$ or $p=\tfrac12$. Only $p=2$ (with $f=-3$) gives a real circle (radius$^2=f^2=9$); $p=\tfrac12$ gives a degenerate point circle. So the centre is $(1,3)$.

Hyperbola $\dfrac{x^2}1-\dfrac{y^2}3=1$; tangent $y=mx\pm\sqrt{m^2-3}$ through $(1,3)$:
$$3-m=\pm\sqrt{m^2-3}\ \Longrightarrow\ (3-m)^2=m^2-3\ \Longrightarrow\ m=2.$$

Solutions · 1