The length of the chord intercepted by the ci | vedclass

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Question

(A) The equation of the circle is $x^2+y^2-6x+8y-5=0$. Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=-3, f=4, c=-5$. Centre $C = (-g, -f) = (3, -4)$.

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(A) The equation of the circle is $x^2+y^2-6x+8y-5=0$. Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=-3, f=4, c=-5$. Centre $C = (-g, -f) = (3, -4)$. Radius $r = \sqrt{g^2+f^2-c} = \sqrt{(-3)^2+4^2-(-5)} = \sqrt{9+16+5} = \sqrt{30}$. The line equation is $2x-y-5=0$. The perpendicular distance $d$ from the centre $(3, -4)$ to the line is: $d = \frac{|2(3)-(-4)-5|}{\sqrt{2^2+(-1)^2}} = \frac{|6+4-5|}{\sqrt{5}} = \frac{5}{\sqrt{5}} = \sqrt{5}$. The length of the chord is given by $2\sqrt{r^2-d^2}$. Length $= 2\sqrt{(\sqrt{30})^2-(\sqrt{5})^2} = 2\sqrt{30-5} = 2\sqrt{25} = 2 	imes 5 = 10$ units. Thus,the correct option is $A$.

The length of the chord intercepted by the circle $x^2+y^2-6x+8y-5=0$ on the line $2x-y=5$ is equal to $L$ units. Find $L$.

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