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JEE MATHEMATICS MCQ: SYSTEM OF CIRCLES

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Given that circle x2+y2−2  x+6  y+6=0{x^2} + {y^2} - 2\;x + 6\;y + 6 = 0 and x2+y2−5  x+6  y+15=0{x^2} + {y^2} - 5\;x + 6\;y + 15 = 0 touch, the equation to their common tangent is
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Correct option: A
x=3x = 3
Option A (x=3x = 3) is correct because the equation of the common tangent is found by subtracting the equations of the two circles, S1−S2=0{S_1} - {S_2} = 0.

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x=1x = 1 is the radical axis of the two orthogonally intersecting circles. If x2+y2=4{x^2} + {y^2} = 4 is one of the circles, then the other circle is The circle x2+y2+8y−4=0{x^2} + {y^2} + 8y - 4 = 0, cuts the real circle x2+y2+gx+4=0{x^2} + {y^2} + gx + 4 = 0 orthogonally, if The equation of the circle whose diameter is the common chord of the circles x2+y2+3x+2y+1=0x^2+y^2+3 x+2 y+1=0 and x2+y2+3x+4y+2=0x^2+y^2+3 x+4 y+2=0, is If the circles x2+y2+2ky+2x+6=0{x^2} + {y^2} + 2ky + 2x + 6 = 0 and x2+y2+2ky+k=0{x^2} + {y^2} + 2ky + k = 0 intersects orthogonally. Then, the value of kk is If the circles x2+y2+2ax+cy+a=0x^2+y^2+2 a x+c y+a=0 and x2+y2−3ax+dy−1=0x^2+y^2-3 a x+d y-1=0 intersect in two distinct points PP and QQ, then the line 5x+by−a=05 x+b y-a=0 passes through PP and QQ for The equation of the circle having its centre on the line x+2y−3=0x + 2y - 3 = 0 and passing through the point of intersection of the circles x2+y2−4y+1=0{x^2} + {y^2} - 4y + 1 = 0 and x2+y2−4x−2y+4=0{x^2} + {y^2} - 4x - 2y + 4 = 0 is
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