Question

A variable circle cuts each of the circles x² + y² - 2x = 0 and x² + y² - 4x - 5 = 0 orthogonally. The variable circle passes through two fixed points whose co-ordinates are:

• A. ($\frac{-5 \pm \sqrt{3}}{2}$,0)
• B. ($\frac{-5 \pm 3\sqrt{5}}{2}$,0)
• C. ($\frac{-5 \pm 5\sqrt{3}}{2}$,0)
• D. ($\frac{-5 \pm \sqrt{5}}{2}$,0)

Answer 1

Answer: (B)

($\frac{-5 \pm 3\sqrt{5}}{2}$,0)

Answer 2

The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$. The condition for two circles to be orthogonal is $2gg' + 2ff' = c + c'$. Applying this to the given circles $x^2 + y^2 - 2x = 0$ ($g'=-1, f'=0, c'=0$) and $x^2 + y^2 - 4x - 5 = 0$ ($g'=-2, f'=0, c'=-5$) yields $-2g=c$ and $-4g=c-5$. Solving these gives $g=5/2$ and $c=-5$. The variable circle's equation becomes $x^2 + y^2 + 5x + 2fy - 5 = 0$. Rearranging this as $(x^2 + y^2 + 5x - 5) + f(2y) = 0$, for it to hold for all $f$, we must have $2y=0$ and $x^2 + y^2 + 5x - 5 = 0$. This leads to $y=0$ and $x^2 + 5x - 5 = 0$. Solving the quadratic for $x$ gives $x = \frac{-5 \pm 3\sqrt{5}}{2}$. Thus, the fixed points are $(\frac{-5 \pm 3\sqrt{5}}{2}, 0)$.