JEE MATHEMATICS MCQ: CIRCLES

1 Easy

1 of 1

Application +4 | -1

The circle whose equations are

{x^2} + {y^2} + {c^2} = 2ax

and

{x^2} + {y^2} + {c^2} - 2by = 0

will touch one another externally if

View answer and explanation

Correct option: C

\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} = \frac{1}{{{c^2}}}

The correct answer is C (

\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} = \frac{1}{{{c^2}}}

) because it represents the condition for the two circles to touch each other externally, derived from the distance between their centers and the sum of their radii.

More Questions from CIRCLES

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The sum of the coefficients in the expansion of

{\left( {{\alpha ^2}{x^2} - 2\alpha \mathrm{ }x + 1} \right)^{51}},

as a polynomial in

x

, vanishes. Position of the point

\left( {\alpha ,2{\alpha ^2}} \right)

with respect to the circle

{x^2} + {y^2} = 4,

is One of the diameter of the circle

{x^2} + {y^2} - 2x + 4y - 4 = 0

is The locus of the middle point of the chords of the circle

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

such that the chords pass through a given point

\left( {{x_1},{y_1}} \right)

, is

\quad \alpha, \beta

and

\gamma

are the parametric angles of three points

P, Q

, and R respectively, on the circle

x^2+y^2=1

, and

A

is the point

(-1,0)

. If the lengths of the chords

A P, A Q

and

A R

are in

G P

, then

\cos \alpha / 2, \cos \beta / 2

and

\cos \gamma / 2

are in The radius of the circle passing through the point (6,2), two of whose diameters are

x+y=6

and

x+2 y=4

is If a circle has two of its diameters along the lines

x+y=5

and

x-y=1

and has area

9 \pi

, then the equation of the circle is