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 $L = 2\sqrt{r^2 - d^2}$,where $r$ is the radius and $d$ is the perpendicular distance from the center $(0, 0)$

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 $L = 2\sqrt{r^2 - d^2}$,where $r$ is the radius and $d$ is the perpendicular distance from the center $(0, 0)$ to the line $\frac{x}{a} + \frac{y}{b} - 1 = 0$.The 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 $d$ into the formula: $L = 2\sqrt{r^2 - (\frac{ab}{\sqrt{a^2 + b^2}})^2} = 2\sqrt{r^2 - \frac{a^2b^2}{a^2 + b^2}}$.Simplifying,we get $L = 2\sqrt{\frac{r^2(a^2 + b^2) - a^2b^2}{a^2 + b^2}}$.

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

Three immiscible liquids of densities $d_1 > d_2 > d_3$ and refractive indices $\mu_1 > \mu_2 > \mu_3$ are put in a beaker. The height of each liquid column is $h/3$. $A$ dot is made at the bottom of the beaker. For near normal vision,find the apparent depth of the dot.

Which one of the following elements,when present as an impurity in silicon,makes it a $p$-type semiconductor?

The relation between the current $i$ (in ampere) in a conductor and the time $t$ (in second) is $i=12t+9t^2$. The charge passing through the conductor between the times $t=2 \ s$ and $t=10 \ s$ is (in $C$)

Which one of the following is used in the preparation of styrene?

Vedclass Test Series

Exam Paper Generator

Online Exam Module