0
$\begingroup$

If the circle $x^2 +y^2 +2gx+2fy+c=0$ cuts the three circles

\begin{align}x^2+y^2-5&=0\\ x^2+y^2-8x-6y+10&=0\\ x^2+y^2-4x+2y-2&=0\end{align}

at the extremities of their diameters, then :

A) $c= -5$

B) $fg=\dfrac{147}{25}$

C) $g+2f=c+2$

D) $4f=3g$

Please explain how to do it. The correct options are A,B,D

$\endgroup$
1
$\begingroup$

The idea here is that if you have two circles $f_1(x, y) = 0$ and $f_2(x, y) = 0$ then $f_1(x, y) - f_2 (x, y) = 0 $ represents the common chord (provided that the coefficients of $x^2$ and $y^2$ are the same in both $f_1$ and $f_2$). And you're given that the common chords are diameters, so they must pass through the centers of all respective circles. You'll get three such equations and you can thus solve for the three variables $g, f, c$.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.