JEE Advanced Mathematics Question (2021) — Solution

Step-by-step solution

Let  A  is the point of intersection of the given lines  y = 0 ,   x + y + 1 = 0 Their point of intersection  A  is  - 1 , 0 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  1 , - 2 Let the third vertex is  h , k Slope of the line perpendicular to  x + y + 1 = 0  is  1 So, the perpendicular from  h , k  to line  x + y + 1 = 0  has slope  1  and passes through the orthocenter  1 , 1 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 + 2 g x + 2 f y = 0 ( 0 , 0 ) → c = 0 ( - 1 , 0 ) → 1 - 2 g = 0 ⇒ g = 1 2 ( 1 , - 2 ) → 5 + 1 - 4 f = 0 ⇒ f = 3 2 Hence, the equation of the circumcircle is  x 2 + y 2 + x + 3 y = 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 1 , 1  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 + 2 g x + 2 f y + c = 0 All the three points  ( 1 , - 1 ) ,   - 2 , - 2   &   - 1 , 0  lies on this circle ( 1 , - 1 ) → 2 + 2 g - 2 f + c = 0   . . . i ( - 1 , 0 ) → 1 - 2 g + c = 0   . . . i i ( - 2 , - 2 ) → 8 - 4 g - 4 f + c = 0   . . . i i i By using the equations  i ,   i i   &   i i i , we get g = 1 2 ,   f = 3 2 ,   c = 0 Hence, the equation of the circle is  x 2 + y 2 + x + 3 y = 0