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Circle question

2008 · Shift 0 · Q46
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Circle question

2008 · Shift 0 · Q46

JEE MainMathematicsCircleMCQ+4 / −1
The point diametrically opposite to the point P(1,0)P(1, 0)P(1,0) on the circle x2+y2+2x+4y−3=0{x^2} + {y^2} + 2x + 4y - 3 = 0x2+y2+2x+4y−3=0 is :
  1. A
    (3,−4)(3, -4)(3,−4)
  2. B
    (−3,4)(-3, 4)(−3,4)
  3. C
    (−3,−4)(-3, -4)(−3,−4)
  4. D
    (3,4)(3, 4)(3,4)
View written solutionFree

Correct answer: C

  1. Write the circle in standard form

Given circle: x2+y2+2x+4y−3=0x^2+y^2+2x+4y-3=0x2+y2+2x+4y−3=0

Complete the squares: x2+2x+y2+4y=3x^2+2x+y^2+4y=3x2+2x+y2+4y=3 x2+2x+1+y2+4y+4=3+1+4x^2+2x+1+y^2+4y+4=3+1+4x2+2x+1+y2+4y+4=3+1+4 (x+1)2+(y+2)2=8(x+1)^2+(y+2)^2=8(x+1)2+(y+2)2=8

So, the center of the circle is: C(−1,−2)C(-1,-2)C(−1,−2)

  1. Use the fact about diametrically opposite points

If P(1,0)P(1,0)P(1,0) and Q(x,y)Q(x,y)Q(x,y) are endpoints of a diameter, then the center is the midpoint of PQPQPQ.

So, (1+x2,0+y2)=(−1,−2)\left(\frac{1+x}{2},\frac{0+y}{2}\right)=(-1,-2)(21+x​,20+y​)=(−1,−2)

  1. Solve for xxx and yyy

For the xxx-coordinate: 1+x2=−1\frac{1+x}{2}=-121+x​=−1 1+x=−21+x=-21+x=−2 x=−3x=-3x=−3

For the yyy-coordinate: y2=−2\frac{y}{2}=-22y​=−2 y=−4y=-4y=−4

Thus, Q=(−3,−4)Q=(-3,-4)Q=(−3,−4)

  1. Match with the options

The correct option is: (−3,−4)\boxed{(-3,-4)}(−3,−4)​ which is Option C.

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