Let the straight line y=2x touch a circle wit | vedclass

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(C) The center of the circle is $P(0, \alpha)$ and the radius is $r$. The line $2x - y = 0$ is tangent to the circle at $A_1$.The perpendicular distan

Vedclass · Accepted Answer

$(P) \rightarrow (4), (Q) \rightarrow (2), (R) \rightarrow (5), (S) \rightarrow (3)$

Vedclass · Answer

(C) The center of the circle is $P(0, \alpha)$ and the radius is $r$. The line $2x - y = 0$ is tangent to the circle at $A_1$.The perpendicular distance from the center $(0, \alpha)$ to the line $2x - y = 0$ is equal to the radius $r$:$\frac{|2(0) - \alpha|}{\sqrt{2^2 + (-1)^2}} = r \implies \frac{|-\alpha|}{\sqrt{5}} = r \implies \alpha = r\sqrt{5}$.Given $\alpha + r = 5 + \sqrt{5}$,substitute $\alpha = r\sqrt{5}$:$r\sqrt{5} + r = 5 + \sqrt{5} \implies r(\sqrt{5} + 1) = \sqrt{5}(\sqrt{5} + 1) \implies r = \sqrt{5}$.Then $\alpha = \sqrt{5} \cdot \sqrt{5} = 5$. So,the center is $P(0, 5)$.The point $A_1$ is the foot of the perpendicular from $P(0, 5)$ to the line $2x - y = 0$:$\frac{x - 0}{2} = \frac{y - 5}{-1} = -\frac{2(0) - 5}{2^2 + (-1)^2} = \frac{5}{5} = 1$.$x = 2(1) = 2$ and $y - 5 = -1 \implies y = 4$. Thus,$A_1 = (2, 4)$.Since $A_1 B_1$ is a diameter,the center $P(0, 5)$ is the midpoint of $A_1 B_1$. Let $B_1 = (x_1, y_1)$:$\frac{x_1 + 2}{2} = 0 \implies x_1 = -2$.$\frac{y_1 + 4}{2} = 5 \implies y_1 + 4 = 10 \implies y_1 = 6$.Thus,$B_1 = (-2, 6)$.Matching the results:$(P) \alpha = 5 \rightarrow (4)$$(Q) r = \sqrt{5} \rightarrow (2)$$(R) A_1 = (2, 4) \rightarrow (5)$$(S) B_1 = (-2, 6) \rightarrow (3)$Therefore,$(P) \rightarrow (4), (Q) \rightarrow (2), (R) \rightarrow (5), (S) \rightarrow (3)$.

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

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