The length of the chord of contact of tangent | vedclass

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(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

$3$

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(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 $\sqrt{2}$)

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