Medium MCQ +4 / -1 PYQ · JEE Mains 2020

The centre of the circle passing through the point (0, 1) and touching the parabola
y = x2 at the point (2, 4) is :

  1. A $\left( {{6 \over 5},{{53} \over {10}}} \right)$
  2. B $\left( {{3 \over {10}},{{16} \over 5}} \right)$
  3. C $\left( {{{ - 53} \over {10}},{{16} \over 5}} \right)$
  4. D $\left( {{{ - 16} \over 5},{{53} \over {10}}} \right)$ Correct answer

Solution

Circle passes through A(0, 1) and B(2, 4). <br><br>y = x<sup>2</sup> <br><br>$\Rightarrow$ ${\left. {{{dy} \over {dx}}} \right|_B}$ = 4 <br><br>tangent at (2,4) is <br><br>(y – 4) = 4(x – 2) <br><br>4x – y – 4 = 0 <br><br>Equation of circle <br><br>(x - 2)<sup>2</sup> + (y–4)<sup>2</sup> + $\lambda$(4x–y - 4) = 0 <br><br>Passing through (0,1) <br><br>$\therefore$ 4 + 9 + $\lambda$(–5) = 0 <br><br>$\Rightarrow$ $\lambda$ = ${{13} \over 5}$ <br><br>$\therefore$ Circle is <br><br>x<sup>2</sup>– 4x + 4 + y<sup>2</sup> – 8y + 16 + ${{13} \over 5}$[4x - y - 4] = 0 <br><br>$\Rightarrow$ x<sup>2</sup> + y<sup>2</sup> + $\left( {{{52} \over 5} - 4} \right)$x - $\left( {8 + {{13} \over 5}} \right)$y + 20 - ${{{52} \over 5}}$ = 0 <br><br>$\Rightarrow$ x<sup>2</sup> + y<sup>2</sup> + ${{{32} \over 5}x - {{53} \over 5}y}$ + ${{48} \over 5}$ = 0 <br><br>$\therefore$ Centre is $\left( {{{ - 16} \over 5},{{53} \over {10}}} \right)$

About this question

Subject: Mathematics · Chapter: Conic Sections · Topic: Parabola

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