Question

Let AB be any chord of the circle $x^2+y^2-4x-4y+4=0$ which subtends an angle of ${90}^0$ at the point $(2,3)$, then the locus of the mid point of AB is circle whose centre is

• A. $(1,5)$
• B. $(1,\frac{5}{2})$
• C. $(1,\frac{3}{2})$
• D. $(2,\frac{5}{2})$

Answer 1

The correct option is D $(2,\frac{5}{2})$

Given circle $S:(x–2{)}^2+(y–2{)}^2=2^2,C=(2,2),r=2$

Let $D=(2,3)$, D lies inside the circle and $E=(x,y)$ be the midpoint of AB

$△CEB$ is right triangle at $E$

$⟹C{B}^2=B{E}^2+E{C}^2$

$⟹B{E}^2=r^2–(\sqrt{(x–2{)}^2+(y–2{)}^2}{)}^2$

$E{B}^2=4–\left(\right(x–2{)}^2+(y-2{)}^2)-\left(1\right)$

Since $AB$ subtends${90}^{\circ }$ at $D$

$\angle ADB={90}^{\circ }$

$D$ lies on a circle whose diameter is $AB$ and $\angle ADB$ is angle in semicircle.

$⟹ED=EB=radius$

$EB+ED=\sqrt{(x–2{)}^2+(y–3{)}^2}$

Substituting $$EB$ value in (1)

$(x–2{)}^2+(y–3{)}^2=4–(x-2{)}^2–(y–2{)}^2$

$⟹2x^2–8x+2y^2–10y+17=0$

Locus of midpoint of $E$ is

$x^2+y^2–4x–5y+\frac{17}{2}=0$ which is a circle centered at $(2,\frac{5}{2})$