JEE MATHEMATICS MCQ: CIRCLES

1 Easy

1 of 1

Application +4 | -1

The equation of a circle which has a tangent

3 x+4 y=6

and two normals given by

(x-1)(y-2)=0

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because the circle's center lies at the intersection of the normals and the radius is the distance from the center to the tangent.

More Questions from CIRCLES

Practice All Chapter MCQs →

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

x

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

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

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

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

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

\quad \alpha, \beta

\gamma

P, Q

x^2+y^2=1

A

(-1,0)

A P, A Q

A R

G P

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

\cos \gamma / 2

x+y=6

x+2 y=4

x+y=5

x-y=1

9 \pi