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JEE Main 20146 April
Exam
JEE Main
Year
2014
Date
6 April
Subject
Mathematics
Chapter
Circle

JEE Main 2014 · 6 April · Mathematics · Circle

Let C be the circle with center at (1, 1) and radius = 1 . If T is...

Let CC be the circle with center at \left(1, 1\right)andradiusand radius= 1.If. IfTisthecirclecenteredat(0,y)is the circle centered at \left(0, y\right), passing through the origin and touching the circle CC externally, then the radius of TT is equal toLet C be the circle with center at (1, 1) and radius = 1 . If T is the circle centered at (0, y) , passing through the

Options

  1. A12\dfrac{1}{2}
  2. B14\dfrac{1}{4}
  3. C32\dfrac{\sqrt{3}}{\sqrt{2}}
  4. D32\dfrac{\sqrt{3}}{2}

Correct answer

B. 14\dfrac{1}{4}

Step-by-step solution

Let rr be the required radius
As origin lies on the circle, the distance of (0,0)\left(0, 0\right) from \left(0, y\right) = r</p><p></p><p>∴ \text{y} = \text{r}<br/><br/><imgsrc="https://cdn−question−pool.getmarks.app/pyq/jeemain/ca900165−d006−4374−a67d−bcde7d10e525−image.png"style="width:230px;height:187px;"/><br/><br/>Asthetwocirclestouchexternally,distancebetweencentresissumoftheradii<br/><br/><br /><br /><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/ca900165-d006-4374-a67d-bcde7d10e525-image.png" style="width: 230px; height: 187px;" /><br /><br />As the two circles touch externally, distance between centres is sum of the radii<br /><br />⇒ \sqrt{{\left(0 - 1\right)}^{2} + {\left(y - 1\right)}^{2}} = r + 1<br/><br/><br /><br />⇒ 1 + {y}^{2} - 2y + 1 = {r}^{2} + 2r + 1<br/><br/><br /><br />⇒ 1 + {r}^{2} - 2r = {r}^{2} + 2r,[∵y=r], \left[∵ y = r\right]

∴1=4r∴ 1 = 4r

∴r=14∴ r = \dfrac{1}{4}

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