JEE MATHEMATICS MCQ: CIRCLES

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Application +4 | -1

If a conic passes through

(1,0)

and satisfies differential equation

\left(1+y^2\right) d x-x y d y=0

. Then the equation of circle which is touching this conic at

(\sqrt{2}, 1)

and passing through one of its foci is:

View answer and explanation

Correct option: A

x^2+y^2-3 \sqrt{2} x-y+4=0

The correct answer is

x^2+y^2-3 \sqrt{2} x-y+4=0

because it represents the equation of the circle that touches the conic at the given point and passes through one of its foci.

More Questions from CIRCLES

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{\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