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Circle question

2022 · 26 Jul · Shift 2 · Q29
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  5. /2022 · 26 Jul · Shift 2 · Q29

Circle question

2022 · 26 Jul · Shift 2 · Q29

JEE MainMathematicsCircleMCQ+4 / −1
Let the abscissae of the two points PPP and QQQ on a circle be the roots of x2−4x−6=0x^{2}-4 x-6=0x2−4x−6=0 and the ordinates of P\mathrm{P}P and Q\mathrm{Q}Q be the roots of y2+2y−7=0y^{2}+2 y-7=0y2+2y−7=0. If PQ\mathrm{PQ}PQ is a diameter of the circle x2+y2+2ax+2by+c=0x^{2}+y^{2}+2 a x+2 b y+c=0x2+y2+2ax+2by+c=0, then the value of (a+b−c)(a+b-c)(a+b−c) is ‾\underline{\hspace{2cm}}​.
  1. A
    12
  2. B
    13
  3. C
    14
  4. D
    16
View written solutionFree

Correct answer: A

  1. Let the two points be P(x1,y1)P(x_1,y_1)P(x1​,y1​) and Q(x2,y2)Q(x_2,y_2)Q(x2​,y2​).

    Given:

    • Abscissae are roots of x2−4x−6=0x^2-4x-6=0x2−4x−6=0
    • Ordinates are roots of y2+2y−7=0y^2+2y-7=0y2+2y−7=0

    So,

    \qquad x_1x_2=-6$$ and $$y_1+y_2=-2, \qquad y_1y_2=-7.$$

  2. Since PPP and QQQ are endpoints of a diameter, the center of the circle is the midpoint of PQPQPQ.

    Hence center is (x1+x22,y1+y22)=(42,−22)=(2,−1).\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)=\left(\frac{4}{2},\frac{-2}{2}\right)=(2,-1).(2x1​+x2​​,2y1​+y2​​)=(24​,2−2​)=(2,−1).

  3. The circle is x2+y2+2ax+2by+c=0.x^2+y^2+2ax+2by+c=0.x2+y2+2ax+2by+c=0. Its center is (−a,−b)(-a,-b)(−a,−b).

    Comparing with center (2,−1)(2,-1)(2,−1), −a=2⇒a=−2,-a=2 \Rightarrow a=-2,−a=2⇒a=−2, −b=−1⇒b=1.-b=-1 \Rightarrow b=1.−b=−1⇒b=1.

  4. Now use the fact that radius squared is r2=PQ24.r^2=\frac{PQ^2}{4}.r2=4PQ2​.

    First compute PQ2PQ^2PQ2: PQ2=(x1−x2)2+(y1−y2)2.PQ^2=(x_1-x_2)^2+(y_1-y_2)^2.PQ2=(x1​−x2​)2+(y1​−y2​)2.

    Now, (x1−x2)2=(x1+x2)2−4x1x2=42−4(−6)=16+24=40. (x_1-x_2)^2=(x_1+x_2)^2-4x_1x_2=4^2-4(-6)=16+24=40.(x1​−x2​)2=(x1​+x2​)2−4x1​x2​=42−4(−6)=16+24=40.

    Similarly, (y1−y2)2=(y1+y2)2−4y1y2=(−2)2−4(−7)=4+28=32. (y_1-y_2)^2=(y_1+y_2)^2-4y_1y_2=(-2)^2-4(-7)=4+28=32.(y1​−y2​)2=(y1​+y2​)2−4y1​y2​=(−2)2−4(−7)=4+28=32.

    Therefore, PQ2=40+32=72.PQ^2=40+32=72.PQ2=40+32=72. So, r2=724=18.r^2=\frac{72}{4}=18.r2=472​=18.

  5. For the circle x2+y2+2ax+2by+c=0,x^2+y^2+2ax+2by+c=0,x2+y2+2ax+2by+c=0, radius satisfies r2=a2+b2−c.r^2=a^2+b^2-c.r2=a2+b2−c.

    Substitute a=−2a=-2a=−2, b=1b=1b=1: 18=(−2)2+12−c=4+1−c=5−c.18=(-2)^2+1^2-c=4+1-c=5-c.18=(−2)2+12−c=4+1−c=5−c. Hence, c=5−18=−13.c=5-18=-13.c=5−18=−13.

  6. Now compute: a+b−c=−2+1−(−13)=12.a+b-c=-2+1-(-13)=12.a+b−c=−2+1−(−13)=12.

  7. Therefore the correct option is: A: 12\boxed{\text{A: }12}A: 12​

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