Question:

Step 1: Set up the problem by identifying what we need to find. Let $M(h, k)$ be the midpoint of a chord of the circle $x^2 + y^2 = 25$ that passes through point $A(1, 2)$. We need to find the locus of all such midpoints. Step 2: Write the equation of a chord given its midpoint. For a circle $x^2 + y^2 = R^2$, the equation of a chord with midpoint $(h, k)$ is: $$hx + ky = h^2 + k^2$$ In our case, with $R^2 = 25$, the chord equation with midpoint $(h, k)$ is: $$hx + ky = h^2 + k^2$$ Step 3: Apply the condition that the chord passes through point $A(1, 2)$. Since the chord must pass through $A(1, 2)$, we substitute $x = 1$ and $y = 2$ into the chord equation: $$h(1) + k(2) = h^2 + k^2$$ $$h + 2k = h^2 + k^2$$ Step 4: Rearrange to obtain the locus equation. Rearranging the equation from Step 3: $$h^2 - h + k^2 - 2k = 0$$ Step 5: Complete the square for both variables. Completing the square for $h$ and $k$: $$\left(h - \frac{1}{2}\right)^2 - \frac{1}{4} + (k - 1)^2 - 1 = 0$$ $$\left(h - \frac{1}{2}\right)^2 + (k - 1)^2 = \frac{1}{4} + 1$$ $$\left(h - \frac{1}{2}\right)^2 + (k - 1)^2 = \frac{5}{4}$$ Step 6: Identify the circle parameters and state the final answer. The locus equation $\left(x - \frac{1}{2}\right)^2 + (y - 1)^2 = \frac{5}{4}$ represents a circle with: - Centre: $(a, b) = \left(\frac{1}{2}, 1\right)$ - Radius: $r = \sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2}$ Therefore, the triplet is: $$(a, b, r) = \left(\frac{1}{2}, 1, \frac{\sqrt{5}}{2}\right)$$ This corresponds to **Option 2**.