If the tangent to the circle ${x^2} + {y^2} = {r^2}$ at the point $(a, b)$ meets the coordinate axes at the point $A$ and $B$, and $O$ is the origin, then the area of the triangle $OAB$ is

  • A

    $\frac{{{r^4}}}{{2ab}}$

  • B

    $\frac{{{r^4}}}{{ab}}$

  • C

    $\frac{{{r^2}}}{{2ab}}$

  • D

    $\frac{{{r^2}}}{{ab}}$

Similar Questions

A pair of tangents are drawn to a unit circle with centre at the origin and these tangents intersect at A enclosing an angle of $60^o$. The area enclosed by these tangents and the arc of the circle is

The line $(x - a)\cos \alpha + (y - b)$ $\sin \alpha = r$ will be a tangent to the circle ${(x - a)^2} + {(y - b)^2} = {r^2}$

$\text { Let } $S$ \text { be the circle in the } xy \text {-plane defined by the equation } x ^2+ y ^2=4 \text {. }$

($1$) Let $E_1, E_2$ and $F_1 F_2$ be the chords of $S$ passing through the point $P_0(1,1)$ and parallel to the $x$-axis and the $y$-axis, respectively. Let $G _1 G _2$ be the chord of $S$ passing through $P _0$ and having slope -$1$ . Let the tangents to $S$ at $E_1$ and $E_2$ meet at $E_3$, the tangents to $S$ at $F_1$ and $F_2$ meet at $F_3$, and the tangents to $S$ at $G_1$ and $G_2$ meet at $G_3$. Then, the points $E_3, F_3$, and $G _3$ lie on the curve

$(A)$ $x+y=4$ $(B)$ $(x-4)^2+(y-4)^2=16$ $(C)$ $(x-4)(y-4)=4$ $(D)$ $x y=4$

($2$) Let $P$ be a point on the circle $S$ with both coordinates being positive. Let the tangent to $S$ at $P$ intersect the coordinate axes at the points $M$ and $N$. Then, the mid-point of the line segment MN must lie on the curve

$(A)$ $(x+y)^2=3 x y$ $(B)$ $x^{2 / 3}+y^{2 / 3}=2^{4 / 3}$ $(C)$ $x^2+y^2=2 x y$ $(D)$ $x^2+y^2=x^2 y^2$

Give the answer or quetion ($1$) and ($2$)

  • [IIT 2018]

The equation of the normal to the circle ${x^2} + {y^2} - 2x = 0$ parallel to the line $x + 2y = 3$ is

The equation of the tangent to the circle ${x^2} + {y^2} = {r^2}$ at $(a,b)$ is $ax + by - \lambda = 0$, where $\lambda $ is