The length of the chord joining the points in | vedclass

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

Question

(B) The given line is $\frac{x}{3} + \frac{y}{4} = 1$,which can be written as $4x + 3y - 12 = 0$.The circle is ${x^2} + {y^2} = \frac{169}{25}$,so the

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) The given line is $\frac{x}{3} + \frac{y}{4} = 1$,which can be written as $4x + 3y - 12 = 0$.The circle is ${x^2} + {y^2} = \frac{169}{25}$,so the radius $r = \sqrt{\frac{169}{25}} = \frac{13}{5}$.The perpendicular distance $d$ from the center $(0, 0)$ to the line $4x + 3y - 12 = 0$ is given by $d = \frac{|4(0) + 3(0) - 12|}{\sqrt{4^2 + 3^2}} = \frac{12}{5}$.The length of the chord is $2\sqrt{r^2 - d^2} = 2\sqrt{(\frac{13}{5})^2 - (\frac{12}{5})^2} = 2\sqrt{\frac{169 - 144}{25}} = 2\sqrt{\frac{25}{25}} = 2(1) = 2$.

The length of the chord joining the points in which the straight line $\frac{x}{3} + \frac{y}{4} = 1$ cuts the circle ${x^2} + {y^2} = \frac{169}{25}$ is

Similar Questions

The length of the chord of contact of tangents drawn from the point $(4,4)$ to the circle $x^2 + y^2 - 2x - 2y - 7 = 0$ is: (in $\sqrt{2}$)

If the chord of contact of tangents from a point $A$ to a given circle passes through $B$,then the circle with $AB$ as a diameter will . . . . . .

The locus of the point of intersection of the tangents at the extremities of a chord of the circle $x^2 + y^2 = a^2$ which touches the circle $x^2 + y^2 = 2ax$ is

If the two circles $x^2+y^2-2x-6y+10-r^2=0$ and $x^2+y^2-8x+2y+8=0$ have a common chord of non-zero length,then

If the circles $x^{2}+y^{2}-2x-2y-7=0$ and $x^{2}+y^{2}+4x+2y+k=0$ cut orthogonally,then the length of the common chord of the circles is

Vedclass Test Series

Exam Paper Generator

Online Exam Module