Properties of Triangles 3 Question 7
7. The radius of the circumcircle of the $\triangle P R S$ is
(a) 5
(b) $3 \sqrt{3}$
(c) $3 \sqrt{2}$
(d) $2 \sqrt{3}$
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Answer:
Correct Answer: 7. (b)
Solution:
- Equation of circumcircle of $\triangle P R S$ is
$(x+1)(x-9)+y^{2}+\lambda y=0$
It will pass through $(1,2 \sqrt{2})$, then
$$ -16+8+\lambda \cdot 2 \sqrt{2}=0 $$
$\Rightarrow \quad \lambda=\frac{8}{2 \sqrt{2}}=2 \sqrt{2}$
$\therefore$ Equation of circumcircle is
$$ x^{2}+y^{2}-8 x+2 \sqrt{2} y-9=0 $$
Hence, its radius is $3 \sqrt{3}$.
Alternate Solution
Let $\angle P S R=\theta \Rightarrow \sin \theta=\frac{2 \sqrt{2}}{2 \sqrt{3}}$
$\therefore \quad \sin \theta=\frac{P R}{2 R}$
$\Rightarrow \quad P R=6 \sqrt{2}=2 R \cdot \sin \theta$
$\Rightarrow \quad R=3 \sqrt{3}$
