Point (1,2) lies on the circle
x2+y2+2x+2y−11=0,
because coordinates of point (1,2) satisfy the equation
x2+y2+2x+2y−11=0 Now,
x2+y2−4x−6y−21=0.....(i)
x2+y2+2x+2y−11=0 3x+4y+5=0 From (i) and (iii),
x2+(−)2−4x−6(−)−21=0 ⇒16x2+9x2+30x+25−64x+72x+120−336=0 ⇒25x2+38x−191=0 From (ii) and (iii),
x2+(−)2+2x+2(−)−11=0 ⇒16x2+9x2+30x+25+32x−24x−40−176=0 ⇒25x2+38x−191=0...(v) Thus we get the same equation
from (ii) and (iii) as we get from equation (i) and(iii).
Hence the point of intersections of (ii) and (iii) will be same as the point of intersections of (i) and (iii).
Therefore the circle (ii) passing through the point of intersection of circle(i) and point (1,2) also as shown in the figure.
Hence equation(ii) i.e.
x2+y2+2x+2y−11=0 is the equation of required circle.