Q90 / 160 · AP EAMCET 2024 (23 May Shift 1) · Mathematics
If $(1,3)$ is the midpoint of a chord of the circle, $x^2+y^2-4 x-8 y+16=0$, then the area of the triangle formed by that chord with the coordinate axes is
- A16
- Correct answerB8
- C4
- D$8 \sqrt{2}$
Solution
Circle : $x^2+y^2-4 x-8 y+16=0$
Equation of chord with given middle point $(1,3)$ is
$\begin{aligned}
& T=S_1 \Rightarrow x+3 y-2(x+1)-4(y+3)+16 \\
& =1+9-4-24+16 \\
& \Rightarrow x+y=4
\end{aligned}$
It cuts co-ordinate axes at $A(0,4)$ and $B(4,0)$ $\triangle A O B$ is right angle. $\therefore \text { Area }=\frac{1}{2} \times O A \times O B=\frac{1}{2} \times 4 \times 4=8$
It cuts co-ordinate axes at $A(0,4)$ and $B(4,0)$ $\triangle A O B$ is right angle. $\therefore \text { Area }=\frac{1}{2} \times O A \times O B=\frac{1}{2} \times 4 \times 4=8$
Hence, the correct answer is option: 'B'.