Let the straight line y = 2x touch a circle with center (0, α), α > 0, and radius r at a point

Question

Let the straight line y = 2x touch a circle with center \((0, \alpha), \alpha > 0\), and radius r at a point \(A_1\). Let \(B_1\) be the point on the circle such that the line segment \(A_1B_1\) is a diameter of the circle. Let \(\alpha +r = 5 + \sqrt {5}.\)

Match each entry in List-I to the correct entry in List-II.

	List-I		List-II
(P)	\(\alpha\) equals	(1)	(-2, 4)
(Q)	r equals	(2)	\(\sqrt {5}\)
(R)	\(A_1\) equals	(3)	(-2, 6)
(S)	\(B_1\) equals	(4)	5
		(5)	(2,4)

The correct option is 

(A) (P) → (4) (Q) → (2) (R) → (1) (S) → (3) 

(B) (P) → (2) (Q) → (4) (R) → (1) (S) → (3) 

(C) (P) → (4) (Q) → (2) (R) → (5) (S) → (3) 

(D) (P) → (2) (Q) → (4) (R) → (3) (S) → (5)

Answer 1

Correct option is (C) (P) → (4) (Q) → (2) (R) → (5) (S) → (3) 

Slope of line = 2 ⇒ \(\tan \theta =2\)

\(c(0,\alpha )\,\alpha > 0\)

\(\alpha+r=5+\sqrt{5}\)  ....(1)

Line \(y = 2x\) is tangent to the circle

\(\therefore\left|\frac{0-\alpha}{\sqrt{4+1}}\right|=r\)

\(\Rightarrow \alpha=r \sqrt{5} \quad \text { as } \alpha>0\)

From equation (1)

\(r \sqrt{5}+r=5+\sqrt{5}\)

\(\Rightarrow r(\sqrt{5}+1)=\sqrt{5}(\sqrt{5}+1) \)

\(\Rightarrow r=\sqrt{5}\)

And \(\alpha=r \sqrt{5}=\sqrt{5} \times \sqrt{5}=5\)

Centre \(C(0,5)\)

\(O C=5 \,\,;A_1 C=\sqrt{5}\)

\(\therefore O A_1=\sqrt{25-5}=\sqrt{20}=2 \sqrt{5} \)

\(\tan \theta=2 \,\, \text { (from figure) }\)

\(\cos \theta=\frac{1}{\sqrt{5}} \quad \sin \theta=\frac{2}{\sqrt{5}} \)

\(A_1\left(0+O A_1 \cos \theta, 0+O A_1 \sin \theta\right) \)

\(A_1\left(2 \sqrt{5} \times \frac{1}{\sqrt{5}}, 2 \sqrt{5} \times \frac{2}{\sqrt{5}}\right) \)

\(A_1(2,4)\)

Let \(B_1\left(x_1, y_1\right)\)

\(\therefore \frac{x_1+2}{2}=0\) and \(\frac{y_1+4}{2}=5\)

\(x_1=-2 , y_1=6 \)

\(B_1(-2,6)\)