Mathematics - Point Question with Solution | TestHub

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}$$(1,-2)\to 5+1-4f=0\Rightarrow f=\frac{3}{2}$Hence, the equation of the circumcircle is$x^2+y^2+x+3y=0$.AliterWe know, the image of the orthocentre about the sides of a triangle lies on circumcircle. The image of$(1,1)$about$y=0$is$(1,-1)$The image of$\left(1,1\right)$about$x+y+1=0$is$(-2,-2)$& intersection of$y=0, x+y+1=0$is$(-1,0).$Let the equation of circle is$x^2+y^2+2gx+2fy+c=0$All the three points$(1,-1), \left(-2,-2\right) \& \left(-1,0\right)$lies on this circle$(1,-1)\to 2+2g-2f+c=0 ...\left(i\right)$$(-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$