Equation of normal to circle {x^2} + {y^2} = 9 at point left( {frac{1}{{√ 2 }},frac{1}{{√ 2 }}}

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10-1.Circle and System of Circles

easy

The equation of the normal to the circle ${x^2} + {y^2} = 9$ at the point $\left( {\frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right)$ is

$x + y = 0$

$x - y = \frac{{\sqrt 2 }}{3}$

$x - y = 0$

None of these

Solution

(c) We know that the equation of normal to the circle ${x^2} + {y^2} = {a^2}$ at the point $({x_1},\;{y_1})$ is $\frac{x}{{{x_1}}} – \frac{y}{{{y_1}}} = 0$.

Therefore , $\frac{x}{{1/\sqrt 2 }} – \frac{y}{{1/\sqrt 2 }} = 0 $

$\Rightarrow x – y = 0$.

Standard 11

Mathematics

The equation of the tangents to the circle ${x^2} + {y^2} + 4x – 4y + 4 = 0$ which make equal intercepts on the positive coordinate axes is given by

medium

If the equation of one tangent to the circle with centre at $(2, -1)$ from the origin is $3x + y = 0$, then the equation of the other tangent through the origin is

hard

The angle at which the circles $(x – 1)^2 + y^2 = 10$ and $x^2 + (y – 2)^2 = 5$ intersect is

normal

If the tangent at the point $P$ on the circle ${x^2} + {y^2} + 6x + 6y = 2$ meets the straight line $5x – 2y + 6 = 0$ at a point $Q$ on the $y$- axis, then the length of $PQ$ is

hard

(IIT-2002)

Let $PQ$ and $RS$ be the tangent at the extremities of the diameter $PR$ of a circle of radius $r$. If $PS$ and $RQ$ intersect at a point $X$ on the circumference of the circle, then $(PQ.RS)$ is equal to