The length of the chord intercepted by the ci | vedclass

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

Question

(B) The length of the chord is given by the formula $L = 2\sqrt{R^2 - d^2}$,where $R$ is the radius of the circle and $d$ is the perpendicular distanc

Vedclass · Accepted Answer

$2\sqrt{\frac{r^2(a^2 + b^2) - a^2b^2}{a^2 + b^2}}$

Vedclass · Answer

(B) The length of the chord is given by the formula $L = 2\sqrt{R^2 - d^2}$,where $R$ is the radius of the circle and $d$ is the perpendicular distance from the center $(0, 0)$ to the line $\frac{x}{a} + \frac{y}{b} - 1 = 0$.The radius $R = r$.The perpendicular distance $d = \frac{|\frac{0}{a} + \frac{0}{b} - 1|}{\sqrt{(\frac{1}{a})^2 + (\frac{1}{b})^2}} = \frac{1}{\sqrt{\frac{a^2 + b^2}{a^2b^2}}} = \frac{|ab|}{\sqrt{a^2 + b^2}}$.Substituting these into the formula:$L = 2\sqrt{r^2 - \frac{a^2b^2}{a^2 + b^2}}$$L = 2\sqrt{\frac{r^2(a^2 + b^2) - a^2b^2}{a^2 + b^2}}$.

The length of the chord intercepted by the circle $x^2 + y^2 = r^2$ on the line $\frac{x}{a} + \frac{y}{b} = 1$ is

Similar Questions

The number of common tangents to the circles $x^2 + y^2 = 1$ and $x^2 + y^2 - 4x + 3 = 0$ is

Let a circle $C$ touch the lines $L_{1}: 4x - 3y + K_{1} = 0$ and $L_{2}: 4x - 3y + K_{2} = 0$,where $K_{1}, K_{2} \in R$. If a line passing through the centre of the circle $C$ intersects $L_{1}$ at $(-1, 2)$ and $L_{2}$ at $(3, -6)$,then the equation of the circle $C$ is:

Suppose that the earth is a sphere of radius $6400 \, km$. The height from the earth's surface from where exactly a fourth of the earth's surface is visible,is $...... \, km$.

In the circle given below,let $OA = 1$ unit,$OB = 13$ units and $PQ \perp OB$. Then,the area of the triangle $PQB$ (in square units) is

The point diametrically opposite to the point $P(1, 0)$ on the circle $x^2 + y^2 + 2x + 4y - 3 = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module