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JEE Main 201511 AprilOnline
Exam
JEE Main
Year
2015
Date
11 April
Mode
Online
Subject
Mathematics
Chapter
Circle

JEE Main 2015 · 11 April · Online · Mathematics · Circle

If a circle passing through the point (- 1, 0) touches y -axis at...

If a circle passing through the point (−1,0)\left(- 1, 0\right) touches yy-axis at \left(0, 2\right),thenthe, then thex$-intercept of the circle isIf a circle passing through the point (- 1, 0) touches y -axis at (0, 2) , then the x -intercept of the circle is

Options

  1. A52\dfrac{5}{2}
  2. B55
  3. C32\dfrac{3}{2}
  4. D33

Correct answer

D. 33

Step-by-step solution

If a circle passing through the point (- 1, 0) touches y -axis at (0, 2) , then the x -intercept of the circle is

Let (a,2)\left(a, 2\right) be the centre of the circle

Equation of the circle is

(x−a)2+(y−2)2=a2{\left(x - a\right)}^{2} + {\left(y - 2\right)}^{2} = {a}^{2}

Put \left(- 1,0\right)intheequationofcirclein the equation of circle{\left(- 1 - a\right)}^{2} + {\left(- 2\right)}^{2} = {a}^{2}<br/><br/><br /><br />1 + 2a + {a}^{2} + 4 = {a}^{2}<br/><br/><br /><br />2a = - 5<br/><br/><br /><br />a = - \dfrac{5}{2}<br/><br/>Equationofcircleis<br/><br/><br /><br />Equation of circle is<br /><br />{\left(x + \dfrac{5}{2}\right)}^{2} + {\left(y - 2\right)}^{2} = \dfrac{25}{4}<br/><br/>xintercept<br /><br />x intercept= 2\sqrt{{g}^{2} - c}<br/><br/><br /><br />= 2\sqrt{\dfrac{25}{4} - 4}<br/><br/><br /><br />= 2\sqrt{\dfrac{9}{4}}<br/><br/><br /><br />= 3
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