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Question

Find the equation of the circle orthogonal to the circles x2+y2+3x5y+6=0 and 4x2+4y 228x+29=0 and whose center lies on the line 3x + 4y + 1 = 0.


A

x2 + y2 + y/4 - 29/4 = 0

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B

2x2 + 2y2 + y - 29 = 0

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C

8x2 + 8y2 + 2y - 29 = 0

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D

4x2 + 4y2 + 2y - 29 = 0

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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+y23x5y+6=0 ............. (2)

4x2+4y228x+29=0

x2+y27x+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

3g5f=c+6 .............(4)

and 2(g)(72)+2(f)(0)=c+294

7g=c+294 ...........(5)

from equation (4) and (5)

3g5f=7g294+6

10g5f=54

2gf=14 .............(6)

centre (-g,-f) lies on the line 3x+4y+1=0

3g4f=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+2y29=0


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