Question

If the circles $${x^2} + {y^2} + 2ax + c = 0$$     and $${x^2} + {y^2} + 2by + c = 0$$     touch each other then :

A. $${a^{ - 2}} + {b^{ - 2}} = {c^{ - 1}}$$  
B. $${a^{ - 2}} + {b^{ - 2}} = {c^{ - 2}}$$
C. $$a + b = 2c$$
D. $$\frac{1}{a} + \frac{1}{b} = \frac{2}{c}$$
Answer :   $${a^{ - 2}} + {b^{ - 2}} = {c^{ - 1}}$$
Solution :
Subtracting the equations, we get $$ax = by.$$   Putting this in the first equation, $${x^2} + {\left( {\frac{a}{b}x} \right)^2} + 2ax + c = 0,$$      which should have equal roots. This gives $$\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} = \frac{1}{c}.$$

Releted MCQ Question on
Geometry >> Circle

Releted Question 1

A square is inscribed in the circle $${x^2} + {y^2} - 2x + 4y + 3 = 0.$$      Its sides are parallel to the coordinate axes. The one vertex of the square is-

A. $$\left( {1 + \sqrt 2 ,\, - 2 } \right)$$
B. $$\left( {1 - \sqrt 2 ,\, - 2 } \right)$$
C. $$\left( {1 - 2 ,\, + \sqrt 2 } \right)$$
D. none of these
Releted Question 2

Two circles $${x^2} + {y^2} = 6$$    and $${x^2} + {y^2} - 6x + 8 = 0$$     are given. Then the equation of the circle through their points of intersection and the point $$\left( {1,\,1} \right)$$  is-

A. $${x^2} + {y^2} - 6x + 4 = 0$$
B. $${x^2} + {y^2} - 3x + 1 = 0$$
C. $${x^2} + {y^2} - 4y + 2 = 0$$
D. none of these
Releted Question 3

The centre of the circle passing through the point (0, 1) and touching the curve $$y = {x^2}$$   at $$\left( {2,\,4} \right)$$  is-

A. $$\left( {\frac{{ - 16}}{5},\,\frac{{27}}{{10}}} \right)$$
B. $$\left( {\frac{{ - 16}}{7},\,\frac{{53}}{{10}}} \right)$$
C. $$\left( {\frac{{ - 16}}{5},\,\frac{{53}}{{10}}} \right)$$
D. none of these
Releted Question 4

The equation of the circle passing through $$\left( {1,\,1} \right)$$  and the points of intersection of $${x^2} + {y^2} + 13x - 3y = 0$$      and $$2{x^2} + 2{y^2} + 4x - 7y - 25 = 0$$      is-

A. $$4{x^2} + 4{y^2} - 30x - 10y - 25 = 0$$
B. $$4{x^2} + 4{y^2} + 30x - 13y - 25 = 0$$
C. $$4{x^2} + 4{y^2} - 17x - 10y + 25 = 0$$
D. none of these

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Circle


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