Question

The circle passing through the points $(1,0),(2,-7)$ and $(8,1)$ also passes through

• A. $(9,6)$
• B. $(9,-6)$
• C. $(0,-3)$
• D. $(0,3)$

Answer 1

Answer: (B), (C)

$(0,-3)$

Let the equation of required circle be
$x^2+y^2+2gx+2fy+c=0$
Above circle is passing through the points $(1,0),(2,-7)$ and $(8,1)$, so
$2g+c=-1\cdots \left(1\right)4g-14f+c=-53\cdots \left(2\right)16g+2f+c=-65\cdots \left(3\right)$

Subtracting equation $\left(1\right)$ from $\left(2\right)$, we get
$2g-14f=-52\Rightarrow g-7f=-26\cdots \left(4\right)$
Subtracting equation $\left(2\right)$ from $\left(3\right)$, we get
$12g+16f=-12\Rightarrow 3g+4f=-3\cdots \left(5\right)$
Using equation $\left(4\right)$ and $\left(5\right)$, we get
$25f=75\Rightarrow f=3$
Using equation $\left(4\right)$, we get
$g=-5$
Using equation $\left(1\right)$, we get
$c=9$

Therefore, the required equation of the circle is
$x^2+y^2-10x+6y+9=0\Rightarrow (x-5{)}^2+(y+3{)}^2=25$
By verification, $(9,-6)$ and $(0,-3)$ lie on the circle.


Alternate method :
Let the vertices of triangle be $A(1,0),B(2,-7)$ and $C(8,1)$
Checking slope of sides
$m_{AB}=\frac{7}{-1}=-7m_{BC}=\frac{-8}{-6}=\frac{4}{3}{m}_{AC}=\frac{-1}{-7}=\frac{1}{7}\Rightarrow m_{AB}\times m_{AC}=-1$
So, it is a right angle triangle at $A(1,0)$
Therefore, $BC$ is a diameter of the circle, so writing circle in diametric form, we get
$(x-2)(x-8)+(y+7)(y-1)=0$
By verification, $(9,-6)$ and $(0,-3)$ lie on the circle.