A line lx + my + n = 0 meets the circle x^2 + | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let the coordinates of point $R$ be $(h, k)$.The equation of the chord of contact of tangents drawn from $R(h, k)$ to the circle $x^2 + y^2 = a^2$

Vedclass · Accepted Answer

$\left( \frac{-a^2l}{n}, \frac{-a^2m}{n} \right)$

Vedclass · Answer

(B) Let the coordinates of point $R$ be $(h, k)$.The equation of the chord of contact of tangents drawn from $R(h, k)$ to the circle $x^2 + y^2 = a^2$ is given by $hx + ky = a^2$,or $hx + ky - a^2 = 0$.It is given that this chord of contact is the line $lx + my + n = 0$.Since both equations represent the same line,their coefficients must be proportional:$\frac{h}{l} = \frac{k}{m} = \frac{-a^2}{n}$.From this,we get:$h = \frac{-a^2l}{n}$ and $k = \frac{-a^2m}{n}$.Therefore,the coordinates of $R$ are $\left( \frac{-a^2l}{n}, \frac{-a^2m}{n} \right)$.

$A$ line $lx + my + n = 0$ meets the circle $x^2 + y^2 = a^2$ at the points $P$ and $Q$. The tangents drawn at the points $P$ and $Q$ meet at $R$. Then,the coordinates of $R$ are:

Similar Questions

The length of the common chord of the circles $(x - a)^{2} + y^{2} = c^{2}$ and $x^{2} + (y - b)^{2} = c^{2}$ is .....

Find the distance between the chords of contact of the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ with respect to the points $(0, 0)$ and $(g, f)$.

The two circles $x^2 + y^2 - 2x + 6y + 6 = 0$ and $x^2 + y^2 - 5x + 6y + 15 = 0$ touch each other. The equation of their common tangent is

If $A$ and $B$ are the points of intersection of the circles $x^2+y^2-4x+6y-3=0$ and $x^2+y^2+2x-2y-2=0$,then the distance between $A$ and $B$ is

If the chord of contact of the point $P(1, 1)$ with respect to the circle $S = x^2 + y^2 + 4x + 6y - 3 = 0$ meets the circle $S = 0$ at $A$ and $B$,then the area of $\triangle PAB$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module