Q87 / 160 · TS EAMCET 2024 (09 May Shift 1) · Mathematics
If $(1,1),(-2,2),(2,-2)$ are 3 points on a circle $S$, then the perpendicular distance from the centre of the circle $S$ to the line $3 x-4 y+1=0$ is
- Correct answerA$\frac{1}{2}$
- B1
- C$\frac{23}{10}$
- D2
Solution
Let $A \equiv(1,1), B \equiv(-2,2), C \equiv(2,-2)$
Slope of $A B=\frac{-1}{3}$
Slope of $A C=\frac{-2-1}{2-1}=-3$
Mid point of $A B=\left(\frac{-1}{2}, \frac{3}{2}\right)$
Mid point of $A C=\left(\frac{3}{2}, \frac{-1}{2}\right)$
Equation of $\perp$ bisector of $A B=y-\frac{3}{2}=3\left(x+\frac{1}{2}\right)$
$\Rightarrow y=3 x+3$ $\qquad ....\mathrm{(i)}$
Equation of $\perp$ bisector of $A C=y+\frac{1}{2}=\frac{1}{3}\left(x-\frac{3}{2}\right)$
$\Rightarrow y=\frac{x}{3}-1$ $\qquad ....\mathrm{(ii)}$
Solving (i) and (ii), $x=\frac{-3}{2} \quad y=\frac{-3}{2}$ Centre of circle $=\left(\frac{-3}{2}, \frac{-3}{2}\right)$ 1 distance of centre from $3 x-4 y+1=0$ is $\left|\frac{\frac{-9}{2}+\frac{12}{2}+1}{5}\right|=\frac{1}{2}$.
Hence, the correct answer is option: 'A'.