JEE Advanced · Mathematics · 12. Circle
Paragraph:
A circle \(C\) of radius 1 is inscribed in an equilateral \(\triangle P Q R\). The points of contact of \(C\) with the sides \(P Q, Q R, R P\) are \(D, E, F\) respectively. The line \(P Q\) is given by the equation
\(\sqrt{3} x+y-6=0\) and the point \(D\) is \(\left(\frac{3 \sqrt{3}}{2}, \frac{3}{2}\right)\). Further, it is given that the origin and the centre of \(C\) are on the same side of the line \(P Q\).Question:
The equation of circle \(C\) is
- A
\((x-2 \sqrt{3})^2+(y-1)^2=1\)
- B
\((x-2 \sqrt{3})^2+\left(y+\frac{1}{2}\right)^2=1\)
- C
\((x-\sqrt{3})^2+(y+1)^2=1\)
- D
\((x-\sqrt{3})^2+(y-1)^2=1\)
Answer & Solution
Correct Answer
(D)
\((x-\sqrt{3})^2+(y-1)^2=1\)
Step-by-step Solution
Detailed explanation
Let centre of circle \(C\) be \((h, k)\).
Then, \(\left|\frac{\sqrt{3} h+k-6}{\sqrt{3+1}}\right|=1\)
\[
\Rightarrow \quad \sqrt{3} h+k-6=2,-2
\]
\[
\Rightarrow \quad \sqrt{3} h+k=4
\]
(Rejecting 2 because origin and centre of \(C\) are on the same side of \(P Q\) ) The point \((\sqrt{3}, 1)\) satisfies Eq. (i).
\(\therefore\) Equation of circle \(C\) is \((x-\sqrt{3})^2+(y-1)^2=1\).
Then, \(\left|\frac{\sqrt{3} h+k-6}{\sqrt{3+1}}\right|=1\)
\[
\Rightarrow \quad \sqrt{3} h+k-6=2,-2
\]
\[
\Rightarrow \quad \sqrt{3} h+k=4
\]
(Rejecting 2 because origin and centre of \(C\) are on the same side of \(P Q\) ) The point \((\sqrt{3}, 1)\) satisfies Eq. (i).
\(\therefore\) Equation of circle \(C\) is \((x-\sqrt{3})^2+(y-1)^2=1\).
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