Two circles $x^{2} + y^{2} + 2 g_{1} x + 2 f_{1} y + c_{1} = 0 a n d x^{2} + y^{2} + 2 g_{2} x + 2 f_{2} y + c^{2} = 0$ are said to be orthogonal. Then $2 g_{1} g_{2} + 2 f_{1} f_{2} = c_{1} + c_{2}$

True

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False

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Solution

The correct option is A
True

Given circles $x^{2} + y^{2} + 2 g_{1} x + 2 f_{1} y + c = 0$ .......(1)

$x^{2} + y^{2} + 2 g_{2} x + 2 f_{2} y + c = 0$ ............(2)

Let $C_{1} & C_{2}$ be the centers and $r_{1} & r_{2}$ radii of circle (1) and (2) respectively. $P & Q$ is the point of

intersection of two circles

$C_{1} (- g_{1} - f_{1})$

$C_{2} (- g_{2} - f_{2})$

$r_{1} = \sqrt{g_{1}^{2} + f_{1}^{2} - c_{1}}, r_{2} = \sqrt{g_{2}^{2} + f_{2}^{2} - c_{2}}$

$C_{1} P = r_{1}$

$C_{2} P = r_{2}$

Right angle triangle $C_{1} P C_{2}$

$(c_{1} c_{2})^{2} = r_{1}^{2} + r_{2}^{2}$

$(g_{1} - g_{2})^{2} + (f_{1} - f_{2})^{2} = g_{1}^{2} + f_{1}^{2} - c_{1} + g_{2}^{2} + f_{2}^{2} - c_{2}$

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

So, give statemebt is correct.

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Two circles $x^{2} + y^{2} + 2 g_{1} x + 2 f_{1} y + c_{1} = 0 a n d x^{2} + y^{2} + 2 g_{2} x + 2 f_{2} y + c^{2} = 0$ are said to be orthogonal. Then $2 g_{1} g_{2} + 2 f_{1} f_{2} = c_{1} + c_{2}$

Q. STATEMENT- 1 : The circles

x^{2} + y^{2} - 2 x - 6 y + 3 = 0 and x^{2} + y^{2} - 6 x - 2 y + 9 = 0

cut each other orthogonally.

and

STATEMENT-2 : If two intersecting circles

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

are orthogonal then

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

Q. Prove that the circle through the origin and cutting the circles

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

and

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

orthogonally is

∣ ∣ ∣ ∣ ∣ \begin{matrix} x^{2} + y^{2} & - x & - y c_{1} & g_{1} & f_{1} c_{2} & g_{2} & f_{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0

Q. If the two circles

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

touch each then

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Two circles x2+y2+2g1x+2f1y+c1=0 and x2+y2+2g2x+2f2y+c2=0 are said to be orthogonal. Then 2g1g2+2f1f2=c1+c2

Two circles $x^{2} + y^{2} + 2 g_{1} x + 2 f_{1} y + c_{1} = 0 a n d x^{2} + y^{2} + 2 g_{2} x + 2 f_{2} y + c^{2} = 0$ are said to be orthogonal. Then $2 g_{1} g_{2} + 2 f_{1} f_{2} = c_{1} + c_{2}$