Two circles S1=x2+y2+2 g1 x+2 f1 y+c1=0 and S 2= x 2+ y 2+2 g 2 x +2 f 2 y + c 2=0 cut each other orthogonally, then:

Question

Two circles S1=x2+y2+2 g1 x+2 f1 y+c1=0 and S 2= x 2+ y 2+2 g 2 x +2 f 2 y + c 2=0 cut each other orthogonally, then: (A) 2 g 1 g 2+2 f 1 f 2= c 1+ c 2 (B) 2 g 1 g 2-2 f 1 f 2= c 1+ c 2 (C) 2 g 1 g 2+2 f 1 f 2= c 1- c 2 (D) 2 g 1 g 2-2 f 1 f 2= c 1- c 2

Tardigrade Admin · Accepted Answer

If two circles intersect at right angle i.e. the tangent at their point of intersection are at right angles, then the circles are called orthogonal circles.
The circles

x^{2} + y^{2} + 2 gx + 2 f y + c = 0

and

x^{2} + y^{2} + 2 g_{1} x + 2 f_{1} y + c_{1} = 0

are orthogonal, if:

2 g g_{1} + 2 f f_{1} = c + c_{1}

Thus, in the given question, the condition will be

2 g_{1} g_{2} + 2 f_{1} f_{2} = c_{1} + c_{2}

Q. Two circles $S_{1} = x^{2} + y^{2} + 2 g_{1} x + 2 f_{1} y + c_{1} = 0$ and $S_{2} = x^{2} + y^{2} + 2 g_{2} x + 2 f_{2} y + c_{2} = 0$ cut each other orthogonally, then:

$2 g_{1} g_{2} + 2 f_{1} f_{2} = c_{1} + c_{2}$

$2 g_{1} g_{2} - 2 f_{1} f_{2} = c_{1} + c_{2}$

$2 g_{1} g_{2} + 2 f_{1} f_{2} = c_{1} - c_{2}$

$2 g_{1} g_{2} - 2 f_{1} f_{2} = c_{1} - c_{2}$

Q. Two circles S1​=x2+y2+2g1​x+2f1​y+c1​=0 and S2​=x2+y2+2g2​x+2f2​y+c2​=0 cut each other orthogonally, then:

2g1​g2​+2f1​f2​=c1​+c2​

2g1​g2​−2f1​f2​=c1​+c2​

2g1​g2​+2f1​f2​=c1​−c2​

2g1​g2​−2f1​f2​=c1​−c2​

Q. Two circles $S_{1} = x^{2} + y^{2} + 2 g_{1} x + 2 f_{1} y + c_{1} = 0$ and $S_{2} = x^{2} + y^{2} + 2 g_{2} x + 2 f_{2} y + c_{2} = 0$ cut each other orthogonally, then:

$2 g_{1} g_{2} + 2 f_{1} f_{2} = c_{1} + c_{2}$

$2 g_{1} g_{2} - 2 f_{1} f_{2} = c_{1} + c_{2}$

$2 g_{1} g_{2} + 2 f_{1} f_{2} = c_{1} - c_{2}$

$2 g_{1} g_{2} - 2 f_{1} f_{2} = c_{1} - c_{2}$