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Circle question

2021 · 27 Jul · Shift 1 · Q35
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  5. /2021 · 27 Jul · Shift 1 · Q35

Circle question

2021 · 27 Jul · Shift 1 · Q35

JEE MainMathematicsCircleMCQ+4 / −1
Let P and Q be two distinct points on a circle which has center at C(2, 3) and which passes through origin O. If OC is perpendicular to both the line segments CP and CQ, then the set {P, Q} is equal to :
  1. A
    {(4, 0), (0, 6)}
  2. B
    {(2+22,3−5),(2−22,3+5)}\{ (2 + 2\sqrt 2 ,3 - \sqrt 5 ),(2 - 2\sqrt 2 ,3 + \sqrt 5 )\}{(2+22​,3−5​),(2−22​,3+5​)}
  3. C
    {(2+22,3+5),(2−22,3−5)}\{ (2 + 2\sqrt 2 ,3 + \sqrt 5 ),(2 - 2\sqrt 2 ,3 - \sqrt 5 )\}{(2+22​,3+5​),(2−22​,3−5​)}
  4. D
    {(−-− 1, 5), (5, 1)}
View written solutionFree

Correct answer: D

  1. Find the radius of the circle

The circle has center C(2,3)C(2,3)C(2,3) and passes through the origin O(0,0)O(0,0)O(0,0). So its radius is

CO=(2−0)2+(3−0)2=4+9=13.CO=\sqrt{(2-0)^2+(3-0)^2}=\sqrt{4+9}=\sqrt{13}.CO=(2−0)2+(3−0)2​=4+9​=13​.

Hence the circle is

(x−2)2+(y−3)2=13.(x-2)^2+(y-3)^2=13.(x−2)2+(y−3)2=13.
  1. Use the condition OC⊥CPOC \perp CPOC⊥CP and OC⊥CQOC \perp CQOC⊥CQ

Vector

OC→=(2,3).\overrightarrow{OC}=(2,3).OC=(2,3).

So any radius vector from CCC to a point P(x,y)P(x,y)P(x,y) on the circle must satisfy

CP→=(x−2,y−3),\overrightarrow{CP}=(x-2, y-3),CP=(x−2,y−3),

and the perpendicularity condition gives

OC→⋅CP→=0.\overrightarrow{OC}\cdot \overrightarrow{CP}=0.OC⋅CP=0.

Thus,

2(x−2)+3(y−3)=02(x-2)+3(y-3)=02(x−2)+3(y−3)=0 2x+3y−13=0.2x+3y-13=0.2x+3y−13=0.

So both PPP and QQQ are points where the line

2x+3y=132x+3y=132x+3y=13

meets the circle

(x−2)2+(y−3)2=13.(x-2)^2+(y-3)^2=13.(x−2)2+(y−3)2=13.
  1. Solve the system

From

2x+3y=13,2x+3y=13,2x+3y=13,

we get

y=13−2x3.y=\frac{13-2x}{3}.y=313−2x​.

Substitute into the circle:

(x−2)2+(13−2x3−3)2=13.(x-2)^2+\left(\frac{13-2x}{3}-3\right)^2=13.(x−2)2+(313−2x​−3)2=13.

Now,

13−2x3−3=13−2x−93=4−2x3=2(2−x)3=−2(x−2)3.\frac{13-2x}{3}-3=\frac{13-2x-9}{3}=\frac{4-2x}{3}=\frac{2(2-x)}{3}=-\frac{2(x-2)}{3}.313−2x​−3=313−2x−9​=34−2x​=32(2−x)​=−32(x−2)​.

So,

(x−2)2+(−2(x−2)3)2=13.(x-2)^2+\left(-\frac{2(x-2)}{3}\right)^2=13.(x−2)2+(−32(x−2)​)2=13. (x−2)2+4(x−2)29=13.(x-2)^2+\frac{4(x-2)^2}{9}=13.(x−2)2+94(x−2)2​=13. 139(x−2)2=13.\frac{13}{9}(x-2)^2=13.913​(x−2)2=13. (x−2)2=9.(x-2)^2=9.(x−2)2=9.

Thus,

x−2=±3  ⟹  x=5 or x=−1.x-2=\pm 3 \implies x=5 \text{ or } x=-1.x−2=±3⟹x=5 or x=−1.

Now find corresponding yyy values using 2x+3y=132x+3y=132x+3y=13:

  • If x=5x=5x=5,

    10+3y=13  ⟹  y=1.10+3y=13 \implies y=1.10+3y=13⟹y=1.

    So one point is (5,1)(5,1)(5,1).

  • If x=−1x=-1x=−1,

    −2+3y=13  ⟹  3y=15  ⟹  y=5.-2+3y=13 \implies 3y=15 \implies y=5.−2+3y=13⟹3y=15⟹y=5.

    So the other point is (−1,5)(-1,5)(−1,5).

Hence,

{P,Q}={(−1,5),(5,1)}.\{P,Q\}=\{(-1,5),(5,1)\}.{P,Q}={(−1,5),(5,1)}.
  1. Check the options

This matches Option D:

{(−1,5),(5,1)}.\{(-1,5),(5,1)\}.{(−1,5),(5,1)}.
  1. Compare with stored correct answer

Stored correct answer: D

Our derived answer: D

So they agree.

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