Consider a triangle Δ whose two sides lie on the x -axis and the line x + y + 1 = 0 . If the…

1. $x^2+y^2-3x+y=0$
2. $x^2+y^2+x+3y=0$
3. $x^2+y^2+2y-1=0$
4. $x^2+y^2+x+y=0$

Solution

Let $A$ is the point of intersection of the given lines $y=0, x+y+1=0$ Their point of intersection $A$ is $\left(-1,0\right)$ Let $B$ is the other point on the line $x+y+1=0$, Now, the perpendicular from $B$ on the $x$-axis will pass through the orthocenter. So, the equation of perpendicular from $B$ on the $x$-axis is $x=1$ Thus, the coordinates of $B$ are $\left(1,-2\right)$ Let the third vertex is $\left(h,k\right)$ Slope of the line perpendicular to $x+y+1=0$ is $1$ So, the perpendicular from $\left(h,k\right)$ to line $x+y+1=0$ has slope $1$ and passes through the orthocenter $\left(1,1\right)$ Thus, its equation is $y=x$ Now, $y=x$ and $x$-axis intersects at the origin. Hence, the vertices of the triangle are $(-1,0), (1,-2) \& (0,0)$ Let the equation of the circle is $x^2+y^2+2gx+2fy=0$ $(0,0)\to c=0$ $(-1,0)\to 1-2g=0\Rightarrow g=\frac{1}{2}$ $(1,-2)\to 5+1-4f=0\Rightarrow f=\frac{3}{2}$ Hence, the equation of the circumcircle is $x^2+y^2+x+3y=0$. Aliter We know, the image of the orthocentre about the sides of a triangle lies on circumcircle. The image of $(1,1)$ about $y=0$ is $(1,-1)$ The image of $\left(1,1\right)$ about $x+y+1=0$ is $(-2,-2)$ & intersection of $y=0, x+y+1=0$ is $(-1,0).$ Let the equation of circle is  $x^2+y^2+2gx+2fy+c=0$ All the three points $(1,-1), \left(-2,-2\right) \& \left(-1,0\right)$ lies on this circle $(1,-1)\to 2+2g-2f+c=0 ...\left(i\right)$ $(-1,0)\to 1-2g+c=0 ...\left(ii\right)$ $(-2,-2)\to 8-4g-4f+c=0 ...\left(iii\right)$ By using the equations $\left(i\right), \left(ii\right) \& \left(iii\right)$, we get $g=\frac{1}{2}, f=\frac{3}{2}, c=0$ Hence, the equation of the circle is $x^2+y^2+x+3y=0$

Asked in: JEE Advanced 2021 (Paper 1)

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