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JEE Advanced 2024Paper 1
Exam
JEE Advanced
Year
2024
Paper
Paper 1
Subject
Mathematics
Chapter
Circle

JEE Advanced 2024 · Paper 1 · Mathematics · Circle

Let the straight line y=2 x touch a circle with center (0, α), α>0...

Let the straight line y=2x\mathrm{y}=2 \mathrm{x} touch a circle with center (0,α),α>0(0, \alpha), \alpha>0, and radius r\mathrm{r} at a point A1\mathrm{A}_1. Let B1\mathrm{B}_1 be the point on the circle such that the line segment A1B1A_1 B_1 is a diameter of the circle. Let α+r=5+5\alpha+r=5+\sqrt{5}. Match each entry in List-I to the correct entry in List-II. Let the straight line y=2 x touch a circle with center (0, α), α>0 , and radius r at a point A_1 . Let B_1 be the point The correct option is

Options

  1. A(P)→(4)(Q)→(2)(R)→(1)(S)→(3)(\mathrm{P}) \rightarrow(4) \quad(\mathrm{Q}) \rightarrow(2) \quad(\mathrm{R}) \rightarrow(1) \quad(\mathrm{S}) \rightarrow(3)
  2. B(P)→(2)(Q)→(4)(R)→(1)(S)→(3)(\mathrm{P}) \rightarrow(2) \quad(\mathrm{Q}) \rightarrow(4) \quad(\mathrm{R}) \rightarrow(1) \quad(\mathrm{S}) \rightarrow(3)
  3. C(P)→(4)(Q)→(2)(R)→(5)(S)→(3)(\mathrm{P}) \rightarrow(4) \quad(\mathrm{Q}) \rightarrow(2) \quad(\mathrm{R}) \rightarrow(5) \quad(\mathrm{S}) \rightarrow(3)
  4. D(P)→(2)(Q)→(4)(R)→(3)(S)→(5)(\mathrm{P}) \rightarrow(2) \quad(\mathrm{Q}) \rightarrow(4) \quad(\mathrm{R}) \rightarrow(3) \quad(\mathrm{S}) \rightarrow(5)

Correct answer

C. (P)→(4)(Q)→(2)(R)→(5)(S)→(3)(\mathrm{P}) \rightarrow(4) \quad(\mathrm{Q}) \rightarrow(2) \quad(\mathrm{R}) \rightarrow(5) \quad(\mathrm{S}) \rightarrow(3)

Step-by-step solution

Let the straight line y=2 x touch a circle with center (0, α), α>0 , and radius r at a point A_1 . Let B_1 be the point Consider centre as P(0,α),α>0∣2(0)−α5∣=r\mathrm{P}(0, \alpha), \alpha>0 \left|\frac{2(0)-\alpha}{\sqrt{5}}\right|=r∣−α∣=5r&α=5r&∴α+r=5+5&5r+r=5(5+1)&r=5,α=5&∴P(0,5)\begin{aligned}& |-\alpha|=\sqrt{5} r \& \alpha=\sqrt{5} r \& \therefore \alpha+r=5+\sqrt{5} \& \sqrt{5} r+r=\sqrt{5}(\sqrt{5}+1) \& r=\sqrt{5}, \alpha=5 \& \therefore P(0,5)\end{aligned}Foot of perpendicular from $\mathrm{P}$ to line $2 \mathrm{x}-\mathrm{y}=0\begin{aligned}& \frac{x-0}{2}=\frac{y-5}{-1}=\frac{-(2(0)-5)}{5}=1 \& x=2, y=4 \quad A_1(2,4)\end{aligned}$$LetLet\mathrm{B}(\mathrm{p}, \mathrm{q}) \therefore \frac{\mathrm{p}+2}{2}=0, \frac{\mathrm{q}+4}{2}=5 \therefore \mathrm{p}=-2, \mathrm{q}=6 \quad \mathrm{~B}(-2,6)$
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