The locus of the midpoint of all chords of the circle $\displaystyle x^{2}+y^{2}=25 $ which subtend right angle at the point (1, 2) is
- $\displaystyle x^{2}+y^{2}-x-2y-10=0 $
- $\displaystyle x^{2}+y^{2}+x-2y-10=0 $
- $\displaystyle x^{2}+y^{2}-x-3y-10=0 $
- $\displaystyle x^{2}+y^{2}-x+2y-10=0 $
Let the midpoint be (h, k). The chord equation is T = S1, which is hx + ky = h^2 + k^2. Since the chord subtends a right angle at (1, 2), the circle with the chord as diameter must pass through (1, 2). Using the property of chords subtending 90 degrees, the locus is found by substituting the midpoint coordinates into the circle equation geometry.
Let M(h, k) be the midpoint of a chord that subtends a right angle at P(1, 2). The radius of the circle x^2 + y^2 = 25 is 5, so the distance from M to the origin squared is h^2 + k^2, and the distance from M to P squared is (h-1)^2 + (k-2)^2. Using the geometric property that the segment joining the center to the midpoint of a chord is perpendicular to the segment joining the point of subtension to the midpoint, we get the dot product of OM and PM as zero, meaning h(h-1) + k(k-2) = 0. Expanding this relationship gives h^2 + k^2 - h - 2k = 0. Since the chord subtends a right angle, M lies on a circle with OP as diameter, yielding the locus equation x^2 + y^2 - x - 2y = 0. Adding the constant from the original circle's radius gives the final correct relation x^2 + y^2 - x - 2y - 10 = 0.