Question

$(-2,3)$ is the middle point of chord $AB$ of the circle $x^2+y^2=81$. The equation of the circle through the points $A,B$ and $(0,1)$ is

• A. $x^2+y^2-16x+24y-23=0$
• B. $x^2+y^2+16x-24y+23=0$
• C. $x^2+y^2-2y+1=0$
• D. $x^2+y^2-16x-24y=0$

Answer 1

Answer: (B)

$x^2+y^2+16x-24y+23=0$

Let $M(-2,3)$ be the mid point of AB. The segment joining the centre of the circle $O$ to $M$ will be perpendicular to AB. Hence, we can write the equation of the segment AB as -
$y-3=\frac{+2}{3}(x+2)$
$3y-2x=13$
Let, the equation of the circle be : $x^2+y^2+2gx+2fy+c=0$
$(0,1)$ lies on the circle. Hence,
$1+2f+c=0$ -----(1)
The common chord of $x^2+y^2=81$ and $x^2+y^2+2gx+2fy+c=0$ is $AB$
We can obtain the equation of $AB$ by subtracting the equations of the two circles.
$2gx+2fy+c+81=0$ -----(3)
We already found that the equation for $AB$ was
$2x-3y+13=0$ -----(4)
The equations (3) & (4) both represent same line. Hence,
$g=\frac{2f}{-3}=\frac{c+81}{13}$
On Solving, we get
$c=23$ $f=-12$ $g=+8$
Hence, the equation of the circle is $x^2+y^2+16x-24y+23=0$
Hence, option 'B' is correct.