Question

If a circle passes through the points $(1,-6),(2,1)$ and $(5,2),$ then the equation of the circle is

• A. $x^2+y^2-10x+4y+29=0$
• B. $x^2+y^2+x+2y+1=0$
• C. $x^2+y^2-14x+6y+16=0$
• D. $x^2+y^2-10x+6y+9=0$

Answer 1

The correct option is D $x^2+y^2-10x+6y+9=0$

Let the equation of the required circle be
$x^2+y^2+2gx+2fy+c=0\cdots \left(i\right)$
According to the problem, the above equation passes through the points $(1,-6),(2,1)$ and $(5,2).$
Therefore, substituting the coordinates of three points $(1,-6),(2,1)$ and $(5,2)$ successively in equation $\left(i\right),$ we get

For the point $(1,-6):$
$1+36+2g-12f+c=0$
$\Rightarrow 2g-12f+c=-37\cdots \left(ii\right)$
For the point $(2,1):$
$4+1+4g+2f+c=0$
$\Rightarrow 4g+2f+c=-5\cdots \left(iii\right)$
For the point $(5,2):$
$25+4+10g+4f+c=0$
$\Rightarrow 10g+4f+c=-29\cdots \left(iv\right)$

Subtracting $\left(ii\right)$ from $\left(iii\right),$ we get
$2g+14f=32$
$\Rightarrow g+7f=16\cdots \left(v\right)$
Again, subtracting $\left(ii\right)$ from $\left(iv\right),$ we get
$8g+16f=8$
$\Rightarrow g+2f=1\cdots \left(vi\right)$
Now, solving equations $\left(v\right)$ and $\left(vi\right),$ we get
$g=-5$ and $f=3.$
Putting the values of $g$ and $f$ in $\left(iii\right),$ we get $c=9.$
Therefore, the equation of the required circle is
$x^2+y^2-10x+6y+9=0$