InterviewSolution
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InterviewSolution
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The equation of the locus of the middle point of a chord of the circle `x^2+y^2=2(x+y)`such that the pair of lines joining the origin to the point ofintersection of the chord and the circle are equally inclined to the x-axisis`x+y=2`(b) `x-y=2``2x-y=1`(d) none of theseA. x+y=2B. `x-y=2`C. `2x-y=1`D. none of these |
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Answer» Correct Answer - A Let (h, k) be the mid-point of a chord of the circle `x^(2)+y^(2)-2x-2y=0`. Then, the equation of the chord is `hx+ky-(x+h) (y | k) = h^(2) |k^(2) 2h 2k` `rArr x(h-1)+y(k-1)=h^(2)+k^(2)-h-k " " ...(i)` The combined equation of the straight lines joining the origin to the points of intersection of the circle and the chord (i) is `x^(2)+y^(2)-2(x+y) {(x(h-1)+y(k-1))/(h^(2)+k^(2)-h-k)}=0` `rArr x^(2)(h^(2)+k^(2)-3h-k+2)+y^(2)(h^(2)+k^(2)-h-3k+2)-2xy(h+k-2)=0` Lines represented by this equation are equally inclined to the x-axis. `:.` Sum of their slopes = 0 `rArr (2(h+k-2))/(h^(2)+k^(2)-h-3k+2)=0 rArr h+k-2=0 rArr h + k =2` Hence, the locus of (h, k) is x+y=2. |
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