Question ID: #566
Let a circle $C$ pass through the points $(4, 2)$ and $(0, 2)$, and its centre lie on $3x+2y+2=0$. Then the length of the chord, of the circle $C$, whose mid-point is $(1, 2)$, is:
- (1) $\sqrt{3}$
- (2) $2\sqrt{3}$
- (3) $4\sqrt{2}$
- (4) $2\sqrt{2}$
Solution:
Let $A(4, 2)$ and $B(0, 2)$ be points on the circle.
Since the $y$-coordinates are equal, the perpendicular bisector of $AB$ is the vertical line $x = \frac{4+0}{2} = 2$.
The centre lies on the intersection of $x=2$ and $3x+2y+2=0$.
$$ 3(2) + 2y + 2 = 0 \Rightarrow 2y = -8 \Rightarrow y = -4 $$
Centre $O \equiv (2, -4)$.
Radius $r = OA = \sqrt{(4-2)^2 + (2-(-4))^2} = \sqrt{4 + 36} = \sqrt{40}$.
Let the given midpoint be $M(1, 2)$. Distance from centre to midpoint:
$$ OM = \sqrt{(2-1)^2 + (-4-2)^2} = \sqrt{1 + 36} = \sqrt{37} $$
Length of chord $= 2\sqrt{r^2 – OM^2}$
$$ = 2\sqrt{40 – 37} = 2\sqrt{3} $$
Ans. (2)
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