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QJEEAdv24Ma642d: JEE Advanced 2024 Mathematics - Class 11 JEEAdv24Ma642d Solution: Let the straight line touch a circle with center , and radius…

Let the straight line y=2xy=2 x touch a circle with center (0, \alpha), \alpha>0 , and radius rr at a point A1A_1 . Let B1B_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.

List-I List-II
(P) α\alpha equals (1) (−2,4)(-2, 4)
(Q) rr equals (2) 5\sqrt{5}
(R) A1A_1 equals (3) (−2,6)(-2, 6)
(S) B1B_1 equals (4) 55
(5) (2,4)(2, 4)

The correct option is

Multiple Choice Options

  • 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)
  • 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)
  • 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)
  • 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)

Step-by-Step Verified Solution

JEE Advanced 2024 Paper 1 Online Mathematics - Circle Question 1 English Explanation

Consider centre as \mathrm{P}(0, \alpha), \alpha>0

∣2(0)−α5∣=r\left|\frac{2(0)-\alpha}{\sqrt{5}}\right|=\mathrm{r}

amp;∣−α∣=5ramp;α=5ramp;∴α+r=5+5amp;5r+r=5(5+1)amp;r=5,α=5amp;∴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 PP to line 2x−y=02 x-y=0

amp;x−02=y−5−1=−(2(0)−5)5=1amp;x=2,y=4A1(2,4)\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}

Let B(p,q)∴p+22=0,q+42=5\mathrm{B}(\mathrm{p}, \mathrm{q}) \quad \therefore \frac{\mathrm{p}+2}{2}=0, \frac{\mathrm{q}+4}{2}=5

∴p=−2,q=6B(−2,6)\therefore \mathrm{p}=-2, \mathrm{q}=6 \quad B(-2,6)

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