The locus of center of the circle which cuts 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 2-0orthogonally isA- The radical axis of the given circleB- A conicC- An ellipseD- Another circle

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Locus

The locus of ...

Question

The locus of center of the circle which cuts 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

An ellipse

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The radical axis of the given circle

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A conic

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Another circle

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Solution

The correct option is B
The radical axis of the given circle

Let required circle be $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$

This circle cuts given two circles orthogonally.

Therefore,

$2 g g_{1} + 2 f f_{1} = c + c_{1}$ - - - - - - (1)

and

$2 g g_{2} + 2 f f_{2} = c + c_{2}$ - - - - - - (2)

Subtracting equation (2) from equation (1)

We get,

$2 g (g_{1} - g_{2}) + 2 f (f_{1} - f_{2}) = c_{1} - c_{2}$ - - - - - - (3)

Centre of the circle is (-g, -f)

While finding the locus of center, we can replace $- g = x & - f = y$ in equation 3

$- 2 x (g_{1} - g_{2}) - 2 y (f_{1} - f_{2}) = c_{1} - c_{2}$

or

$2 x (g_{1} - g_{2}) + 2 y (f_{1} - f_{2}) + c_{1} - c_{2} = 0,$

This is same as $s_{1} - s_{2} = 0$ or radical axis of the given circles

So, option B is correct

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Similar questions

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

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. Assertion :The equation of the circle passing through the point

(2 a, 0)

and whose radical axis is

x = \frac{a}{2}

with respect to the circle

x^{2} + y^{2} = a^{2}

, will be

x^{2} + y^{2} - 2 a x = 0

Reason: The equation of radical axis of two 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

2 (g_{1} - g_{2}) x + 2 (f_{1} - f_{2}) y + (c_{1} - c_{2}) = 0

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. Find the equation of a circle passing through the origin and cutting the circles

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

and

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

orthogonally.

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The locus of center of the circle which cuts the circles x2 + y2 + 2g1x + 2f1y + c1 = 0 and x2 + y2 + 2g2x + 2f2y + c2 = 0 orthogonally is

The locus of center of the circle which cuts 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