At which point on the line x = 3 are the tang | vedclass

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(A) The locus of the point of intersection of perpendicular tangents to a circle is called the director circle. For a circle $x^2 + y^2 = r^2$,the equ

Vedclass · Accepted Answer

$(3, \sqrt{7})$

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(A) The locus of the point of intersection of perpendicular tangents to a circle is called the director circle. For a circle $x^2 + y^2 = r^2$,the equation of the director circle is $x^2 + y^2 = 2r^2$. Here,the given circle is $x^2 + y^2 = 8$,so $r^2 = 8$. The equation of the director circle is $x^2 + y^2 = 2(8) = 16$. We are given that the point lies on the line $x = 3$. Substituting $x = 3$ into the equation of the director circle: $(3)^2 + y^2 = 16$ $9 + y^2 = 16$ $y^2 = 7$ $y = \pm \sqrt{7}$. Thus,the points are $(3, \sqrt{7})$ and $(3, -\sqrt{7})$. Comparing this with the given options,option $A$ is correct.

At which point on the line $x = 3$ are the tangents drawn to the circle $x^2 + y^2 = 8$ perpendicular to each other?

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