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A isoscles triangle is inscribed in the circle \( \Large x^{2}+y^{2}-6x-8y=0 \) with vertex at the origin and one of the equal side along the axis of x. Equation of the other side through the origin is:

A) \( \Large 7x-24y=0 \)
B) \( \Large 24x-7y=0 \)
C) \( \Large 7x+24y=0 \)
D) \( \Large 24x+7y=0 \)

Correct Answer:




D) \( \Large 24x+7y=0 \)

Description for Correct answer:

Centre'of the circle is \( \Large \left(3,\ 4\right) \) and it passes through the origin, If \( \Large y = mx \) is the equation of the required line, then length of' the perpendicular from the centre on this line is equal to the length of the perpendicular from the centre on the axis of x.

= \( \Large \frac{3m-4}{\sqrt{1+m^{2}}}= \pm 4 \)

=> \( \Large 9m^{2}-4m+16 = 16 \left(1+m^{2}\right) \)

=> \( \Large m = -\frac{24}{7} \) (\( \Large \therefore\ m = 0 \) corresponds to x-axis)

and hence, the required equation is \( \Large 24x +\ 7y = 0 \)

Part of solved Circles questions and answers : >> Elementary Mathematics >> Circles

Comments

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