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JEE Main 20249 AprilMorning Shift
Exam
JEE Main
Year
2024
Date
9 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Circle

JEE Main 2024 · 9 April · Morning Shift · Mathematics · Circle

Let the centre of a circle, passing through the points (0,0),(1,0)...

Let the centre of a circle, passing through the points (0,0),(1,0)andtouchingthecircle(0,0),(1,0) and touching the circlex^2+y^2=9,be, be(h, k).Thenforallpossiblevaluesofthecoordinatesofthecentre. Then for all possible values of the coordinates of the centre(h, k), 4(h2+k2)\left(h^2+k^2\right) is equal to_________

Correct answer

0

Step-by-step solution

(x−h)2+(y−k)2=h2+k2x2+y2−2hx−2ky=0∵ passes through (1,0)⇒1+0−2h=0⇒h=1/2∵OC=OP2(12)2+k2=32\begin{aligned} & (\mathrm{x}-\mathrm{h})^2+(\mathrm{y}-\mathrm{k})^2=\mathrm{h}^2+\mathrm{k}^2 \\ & \mathrm{x}^2+\mathrm{y}^2-2 \mathrm{hx}-2 \mathrm{ky}=0 \\ & \because \text{ passes through }(1,0) \\ & \Rightarrow 1+0-2 \mathrm{ h}=0 \\ & \Rightarrow \mathrm{h}=1 / 2 \\ & \because \mathrm{OC}=\frac{\mathrm{OP}}{2} \\ & \sqrt{\left(\frac{1}{2}\right)^2+\mathrm{k}^2}=\frac{3}{2}\end{aligned}14+k2=94k2=2k=±2\begin{aligned} & \frac{1}{4}+\mathrm{k}^2=\frac{9}{4} \\ & \mathrm{k}^2=2 \\ & \mathrm{k}= \pm \sqrt{2} \end{aligned} \thereforePossiblecoordinateofPossible coordinate ofc(h,k)(12,2)(12,−2)4(h2+k2)=4(14+2)=4(94)=9\begin{aligned} & \mathrm{c}(\mathrm{h}, \mathrm{k})\left(\frac{1}{2}, \sqrt{2}\right)\left(\frac{1}{2},-\sqrt{2}\right) \\ & 4\left(\mathrm{ h}^2+\mathrm{k}^2\right)=4\left(\frac{1}{4}+2\right)=4\left(\frac{9}{4}\right)=9 \end{aligned}$
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