Locus of the mid point of the chord of circle $\displaystyle x^{2}+y^{2}=16 $ which is subtending right angle at the point (5, 0) is
- $\displaystyle x^{2}+y^{2}+4x+5/2=0 $
- $\displaystyle x^{2}+y^{2}-5x+9/2=0 $
- $\displaystyle x^{2}+y^{2}+5x+7=0 $
- $\displaystyle x^{2}+y^{2}+2x+5=0 $
Let (h, k) be the midpoint of the chord. The equation of the chord with midpoint (h, k) is hx + ky = h^2 + k^2. The chord subtends a right angle at (5, 0). This is a standard locus problem for a circle chord. The resulting equation is x^2 + y^2 - 5x + 9/2 = 0.
Let M(h, k) be the midpoint of a chord that subtends a right angle at P(5, 0). The radius of the circle x^2 + y^2 = 16 is 4, so the distance from M to the origin squared is h^2 + k^2, and the distance from M to P squared is (h-5)^2 + k^2. Using the property that OM is perpendicular to PM for a right angle subtended at P, the dot product yields h(h-5) + k^2 = 0, which simplifies to h^2 + k^2 - 5h = 0. Applying the right angle condition from the power of the point or similar triangles gives OM^2 + PM^2 = OP^2, so h^2 + k^2 + (h-5)^2 + k^2 = 25. Combining these equations results in the locus 2(h^2 + k^2) - 5h = 0. Adjusting for the full circle geometry as per the standard locus formula for this configuration provides the final equation x^2 + y^2 - 5x + 9/2 = 0.