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JEE MATHEMATICS MCQ: SYSTEM OF CIRCLES

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Equation of a common chord of the circles x2+y2+6x−10y+9=0x^2+y^2+6 x-10 y+9=0 and x2+y2−10x+6y+25=0x^2+y^2-10 x+6 y+25=0 is
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Correct option: D
x−y−1=0x-y-1=0
Option D (x−y−1=0x-y-1=0) is correct because it represents the equation obtained by subtracting the equations of the two circles, which gives the common chord.

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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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