The length of the chord of contact of tangent | vedclass

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

Question

(B) The equation of the circle is $x^2 + y^2 - 2x - 2y - 7 = 0$. The center $C$ is $(1, 1)$ and the radius $r = \sqrt{1^2 + 1^2 - (-7)} = \sqrt{9} = 3

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) The equation of the circle is $x^2 + y^2 - 2x - 2y - 7 = 0$. The center $C$ is $(1, 1)$ and the radius $r = \sqrt{1^2 + 1^2 - (-7)} = \sqrt{9} = 3$.Let $P = (4, 4)$. The distance $PC = \sqrt{(4-1)^2 + (4-1)^2} = \sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2}$.Let $L$ be the length of the tangent from $P$ to the circle. $L = \sqrt{PC^2 - r^2} = \sqrt{(3\sqrt{2})^2 - 3^2} = \sqrt{18 - 9} = \sqrt{9} = 3$.In the right-angled triangle $	riangle PAC$ (where $A$ is the point of contact),the length of the chord of contact $AB$ is given by $2 	imes 	ext{area}(	riangle PAC) / PC$.The area of $	riangle PAC = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes L 	imes r = \frac{1}{2} 	imes 3 	imes 3 = \frac{9}{2}$.Thus,$AB = \frac{2 	imes (9/2)}{3\sqrt{2}} = \frac{9}{3\sqrt{2}} = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2}$.Wait,re-evaluating: The length of the chord of contact is $\frac{2Lr}{\sqrt{L^2 + r^2}}$.Here $L=3, r=3$. So,$AB = \frac{2 	imes 3 	imes 3}{\sqrt{3^2 + 3^2}} = \frac{18}{\sqrt{18}} = \frac{18}{3\sqrt{2}} = \frac{6}{\sqrt{2}} = 3\sqrt{2}$.

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 $\sqrt{2}$)

Similar Questions

The equation of the chord of contact of the circle $x^2 + y^2 + 4x + 6y - 12 = 0$ with respect to the point $(2, 3)$ is:

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

The distance from the centre of the circle $x^2 + y^2 = 2x$ to the straight line passing through the points of intersection of the two circles $x^2 + y^2 + 5x - 8y + 1 = 0$ and $x^2 + y^2 - 3x + 7y - 25 = 0$ is-

If the chord $L \equiv y-mx-1=0$ of the circle $S \equiv x^2+y^2-1=0$ touches the circle $S_1 \equiv x^2+y^2-4x+1=0$,then the possible points for which $L=0$ is a chord of contact of $S=0$ are

The length (in units) of the common chord of the circles $x^2+y^2+2x+3y+1=0$ and $x^2+y^2+4x+3y+2=0$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module