Question

The line $2x+3y=1$ intersects the circle $x^2+y^2=4$ at $A$ and $B$. If the equation of the circle on AB as diameter is $x^2+y^2+2gx+2fy+c=0$ then $c=$

• A. $-50$
• B. $-\frac{54}{13}$
• C. $\frac{50}{13}$
• D. $-\frac{50}{13}$

Answer 1

The correct option is D $-\frac{50}{13}$

To find the intersection points of circle and the line we need to solve their equations simultaneously.

$\because 2x+3y=1\Rightarrow 3y=1-2x\Rightarrow y=\frac{1-2x}{3}$

Putting this value of y in equation of circle, $x^2+(\frac{1-2x}{3}{)}^2=4$

$\Rightarrow 9x^2+1+4x^2-4x=36\Rightarrow 13x^2-4x-35=0$

$\therefore x=\frac{-(-4)\pm \sqrt{(-4{)}^2-4\times 13\times (-35)}}{2\times 13}=\frac{4\pm 6\sqrt{51}}{26}=\frac{2\pm 3\sqrt{51}}{13}$

$\therefore y=\frac{1-2\left[\frac{2\pm 3\sqrt{51}}{13}\right]}{3}\Rightarrow y=\frac{3\mp 2\sqrt{51}}{13}$

Thus the coordinates of A and B are as follows:- $A\equiv (\frac{2+3\sqrt{51}}{13},\frac{3-2\sqrt{51}}{13})$ and $B\equiv (\frac{2-3\sqrt{51}}{13},\frac{3+2\sqrt{51}}{13})$

Now, equation of circle with ends of diameter given is:-$(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$

$\Rightarrow (x-\frac{2+3\sqrt{51}}{13})(x-\frac{2-3\sqrt{51}}{13})+(y-\frac{3-2\sqrt{51}}{13})(y-\frac{3+2\sqrt{51}}{13})=0$

$\Rightarrow x^2-[\frac{2+3\sqrt{51}}{13}+\frac{2-3\sqrt{51}}{13}]x+\left(\frac{2+3\sqrt{51}}{13}\right)\left(\frac{2-3\sqrt{51}}{13}\right)+$

$y^2-[\frac{3-2\sqrt{51}}{13}+\frac{3+2\sqrt{51}}{13}]y+\left(\frac{3-2\sqrt{51}}{13}\right)\left(\frac{3+2\sqrt{51}}{13}\right)=0$

$\Rightarrow x^2-\frac{4}{13}x+\left(\frac{4-9\times 51}{169}\right)+y^2-\frac{6}{13}y+\left(\frac{9-4\times 51}{169}\right)=0$

$\Rightarrow x^2+y^2-\frac{4}{13}x-\frac{6}{13}y+\left(\frac{4+9-(9+4)\times 51}{169}\right)=0$

$\Rightarrow x^2+y^2-\frac{4}{13}x-\frac{6}{13}y-\frac{50}{13}=0$

Thus, $c=\frac{-50}{13}$