JEE MATHEMATICS MCQ: CIRCLES

1 Easy

1 of 1

Application +4 | -1

The equation of circle which touches the line

x+3 y-2=0

and

3 x+9 y-2=0

, also the point of contact on first line is

(-1,1)

, is

View answer and explanation

Correct option: C

x^2+y^2+\frac{32 x}{15}-\frac{8 y}{5}+\frac{26}{15}=0

Option C (

x^2+y^2+\frac{{32 x}}{{15}}-\frac{{8 y}}{{5}}+\frac{{26}}{{15}}=0

) is correct because the circle touches both lines and the point of contact on the first line is given, allowing us to find the other diametric endpoint and thus the circle's equation.

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