The equation of a circle which touches the line $y=x$ at $(1,1)$ and having $y=x-3$ as a normal, is

JEE MATHEMATICS MCQ: CIRCLES

1 Easy

1 of 1

Application +4 | -1

The equation of a circle which touches the line

y=x

(1,1)

and having

y=x-3

as a normal, is

View answer and explanation

Correct option: A

4 x^2+4 y^2-20 x+4 y+8=0

Option A (

4 x^2+4 y^2-20 x+4 y+8=0

) is correct because it represents the equation of a circle that touches the line y=x at (1,1) and has y=x-3 as a normal.

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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