JEE MATHEMATICS MCQ: CIRCLES

1 Easy

1 of 1

Application +4 | -1

Find the radius of the circle whose centre is in fourth quadrant and which touches as each of the three lines

x=0

y=0

and

3 x+4 y-12=0.

View answer and explanation

Correct option: A

Option A (3) is correct because the radius of the circle is found by using the distance from the center to the line and solving the resulting equation.

More Questions from CIRCLES

Practice All Chapter MCQs →

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