Question

Consider a triangle $\Delta$ whose two sides lie on the $x-$axis and the line $x+y+1=0.$ If the orthocenter of $\Delta$ is $(1,1),$ then the equation of the circle passing through the vertices of the triangle $\Delta$ is

• A. $x^2+y^2-3x+y=0$
• B. $x^2+y^2+x+3y=0$
• C. $x^2+y^2+2y-1=0$
• D. $x^2+y^2+x+y=0$

Answer 1

The correct option is B $x^2+y^2+x+3y=0$

As we know mirror image of orthocenter lie on circumcircle.
$\therefore$ Image of $(1,1)$ in $x-$axis is $(1,-1)$

Image of $(1,1)$ in $x+y+1=0$ is given by
$\frac{x-1}{1}=\frac{x-1}{1}=-\frac{2(1+1+1)}{1^2+1^2}$
$\Rightarrow x=-2$ and $y=-2$
$\therefore$ Image of $(1,1)$ in $x+y+1=0$ is $(-2,-2).$
and intersection of $x-$ axis and line $x+y+1=0$ is $(-1,0).$
$\therefore$ The required circle will be passing through the points $(1,–1),(-2,-2)$ and $(-1,0),$
Hence, the equation of circle is $x^2+y^2+x+3y=0$

Alternative solution:
Let the equation of circle
$x^2+y^2+2gx+2fy+c=0$
At $(1,-1):$
$2g-2f+c+2=0\cdots \left(1\right)$
At $(-1,0):$
$-2g+c+1=0\cdots \left(2\right)$
At $(-2,-2):$
$4g+4f-c-8=0\cdots \left(3\right)$
From $\left(1\right),\left(2\right)$ and $\left(3\right),$ we get
$g=\frac{1}{2},f=\frac{3}{2}$ and $c=0$
So, equation of circle is $x^2+y^2+x+3y=0$