The mid point of the chord $x-2y+7=0$ w.r.t the circle $x^2+y^2-2x-10y+1=0$ is?
- $(7, 21)$
- $\left(\dfrac{7}{4}, \dfrac{21}{4}\right)$
- $\left(\dfrac{7}{5}, \dfrac{21}{5}\right)$
- $\left(\dfrac{-7}{5}, \dfrac{-21}{5}\right)$
The center of the circle is (1, 5). The midpoint of a chord is the projection of the center onto the line x - 2y + 7 = 0. The line passing through (1, 5) perpendicular to the chord is 2x + y - 7 = 0; solving the system of these two lines gives (7/5, 21/5).
Let the midpoint of the chord be (h, k). The equation of the circle is x^2 + y^2 - 2x - 10y + 1 = 0, with center at (1, 5). The line segment from the center to the midpoint of a chord is perpendicular to the chord, so the slope of the line connecting (1, 5) and (h, k) is the negative reciprocal of the chord's slope. The chord x - 2y + 7 = 0 has a slope of 1/2, making the required perpendicular slope -2. Setting up the equation (k - 5) / (h - 1) = -2 gives the relation k = -2h + 7. Since (h, k) must lie on the chord's equation, substituting k gives h - 2(-2h + 7) + 7 = 0, which solves to h = 7/5. Substituting h back into the relation gives k = -2(7/5) + 7 = 21/5, making the midpoint (7/5, 21/5).