Q.2. Find the equation of the circle drawn on the common chord of the following pair of circles as diameter x 2 + y 2 - 6 x - 4 y = 12   ;   x 2 + y 2 - 2 x - 6 y = 15 .

Dear Student,

Let S1=x2+y2-6x-4y-12=0 and S2=x2+y2-2x-6y-15=0The equation of common chord is S1-S2=0x2+y2-6x-4y-12-x2+y2-2x-6y-15=0x2+y2-6x-4y-12-x2-y2+2x+6y+15=02x-6x+6y-4y+15-12=02y-4x+3=0Also the equation of circle passing through two circles isS1+λS2=0x2+y2-6x-4y-12+λx2+y2-2x-6y-15=0x2+y2-6x-4y-12+λx2+λy2-2xλ-6yλ-15λ=0x21+λ+y21+λ-2xλ+3-2y3λ+2-34+5λ=0x2+y2-2x×λ+3λ+1-2y×3λ+2λ+1-3×4+5λλ+1=0This is similar to equation of circlex2+y2+2gx+2fy+c=0where centre is -g,-fnow finding value of g and f2g=-2×λ+3λ+1g=-λ+3λ+12f=-2×3λ+2λ+1f=-3λ+2λ+1Hence centre of circle is λ+3λ+1,3λ+2λ+1As this point lie on the common chord above 2y-4x+3=0it will satify the equationputting values in 2y-4x+3=02×3λ+2λ+1-4×λ+3λ+1+3=06λ+4λ+1-4λ+12λ+1+3=06λ+4-4λ-12+3λ+3=05λ-5=0λ=1Hence equation of circle is x21+λ+y21+λ-2xλ+3-2y3λ+2-34+5λ=0putting values we getx21+1+y21+1-2x1+3-2y3+2-34+5=02x2+2y2-8x-10y-27=0
Regards,

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