Question

# Find the locus of the mid point of the chord of a circle $$x^{2} + y^{2} = 4$$ such that the segment intercepted by the chord on the curve $$x^{2} - 2x - 2y = 0$$ subtends a right angle at the origin.

A
x2+y22x2y=0
B
x2+y2+2x2y=0
C
x2+y2+2x+2y=0
D
x2+y22x+2y=0

Solution

## The correct option is B $$x^{2} + y^{2} - 2x - 2y = 0$$Let the midpoint be $$(h,k)$$Equation of chord $$T=S$$$$\Rightarrow hx+ky=h^{2}+k^{2}$$Let the above line intersects curve $$x^{2}-2x-2y=0$$ at $$A$$ and $$B$$.Let $$O$$ be the origin.For finding $$OA \times OB$$, we will use homogenization.$$x^{2}-2x \left(\dfrac{(hx+ky)}{h^{2}+k^{2}}\right)-2y\left(\dfrac{hx+ky}{h^{2}+k^{2}}\right)=0$$$$x^{2}(h^{2}+k^{2})-x(hx+ky)-2y(hx+ky)=0$$$$\Rightarrow x^{2}(h^{2}+k^{2})+x^{2}(-2h)-2kxy-2hxy-2ky^{2}=0$$$$\Rightarrow x^{2}(h^{2}+k^{2}-2h)+y^{2}(-2k)+y(-2hx-2kx)=0$$$$\because OA$$ and $$OB$$ are perpendicular.$$\Rightarrow h^{2}+k^{2}-2h-2k=0$$$$\Rightarrow h^{2}+k^{2}=2(h+k)$$$$\therefore$$ Locus is$$x^{2}+y^{2}=2(x+y)$$$$\Rightarrow x^{2}+y^{2}-2x-2y=0$$Maths

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