Question

# Find the equation of the circle orthogonal to the circles x2+y2+3x−5y+6=0 and 4x2+4y 2−28x+29=0 and whose center lies on the line 3x + 4y + 1 = 0.x2 + y2 + y/4 - 29/4 = 0 2x2 + 2y2 + y - 29 = 0 8x2 + 8y2 + 2y - 29 = 0 4x2 + 4y2 + 2y - 29 = 0

Solution

## The correct option is D 4x2 + 4y2 + 2y - 29 = 0 Let the required circle be x2+y2+2gx+2fy+c=0             .............(1) Given circles are x2+y2−3x−5y+6=0                  ............. (2) 4x2+4y2−28x+29=0                x2+y2−7x+294=0         .............(3) Since circles (1)&(2) and  (1)&(3) are orthogonal to each other then 2g1g2+2f1f2=c1+c2 2(−g)(−32)+2(−f)(+52)=c+6 3g−5f=c+6                               .............(4) and 2(−g)(72)+2(−f)(0)=c+294 −7g=c+294                      ...........(5) from equation (4) and (5) 3g−5f=−7g−294+6 10g−5f=−54 2g−f=14                         .............(6) centre (-g,-f) lies on the line 3x+4y+1=0        −3g−4f=−1                          .............(7) Solving equation (6) and (7) We get, g=0 and f14 From equation (5) c=294 Substituting the values of g, f and c in equation (1) Required equation of circle is x2+y2+2(0)x+2(14)y+(−294)=0 4x2+4y2+2y−29=0

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