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Question

Find the equations of direct common tangents for two circles

x2 + y2 + 6x − 2y + 1 = 0, x2 + y2 − 2x − 6y + 9 = 0


A

35x2 + 12xy 18x = 0

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B

35y2 + 12xy 18y = 0

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C

3x2 4xy + 16y 12x = 0

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D

3y2 4xy + 16x 12y = 0

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Solution

The correct option is D

3y2 4xy + 16x 12y = 0


Given circles,

x2+y2+6x2y+1=0 - - - - - - (1)

x2+y22x6y+9=0 - - - - - - (2)

Let the circle be c1&c2 and radii r1&r2 of circle (1) and (2) respectively

c1(3, 1)

c2(1,3)

r1=g2+f2c=9+11=3

r2=g2+f2c=1+99=1

c1c2=16+4=25=4.47

c1c2>r1+r2

Both the circle lies outside each other.

Let intersection point of direct common tangent be Q. Q divides c1 & c2

externally in the ratio of r1:r2 here, it divides in the ratio of 3:1. Co-ordinates of

Q(x,y)=(m1x2+m2x1m1+m2,m1y2+m2y1m1+m2)

=(3×1+1×1(3)31,3×31×(1)31)

=(62,82)=(3,4)

Equation of direct common tangent or pair of tangent from the external point

T2=SS1

T=xx1+yy1+3(x+x1)(y+y1)+1

=3x+4y+3(x+3)(y+4)+1

=6x+3y+6

Substituting values of T & s1 in equation (3)

(6x+3y+6)2=(x2+y2+6x2y+1) (32+42+188+1)

=(x2+y2+6x2y+1)(36)

36x2+9y2+36+36xy+36y+72x=36x2+36y2+216x72y+36

27y2+36xy144x+108y=0

3y24xy+16x12y=0

Equation of transverse common tangents

3y2+4xy+16x12y=0


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