Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
121 views
in Mathematics by (46.4k points)
closed by

Let the centre of a circle, passing through the point \((0,0),(1,0)\) and touching the circle \(x^{2}+y^{2}=9\), be (h, k). Then for all possible values of the coordinates of the centre \( (h, k), 4\left(h^{2}+k^{2}\right)\) is equal to ______.

1 Answer

+1 vote
by (49.7k points)
selected by
 
Best answer

Correct answer is :

Let the centre of a circle

\((x-h)^{2}+(y-k)^{2}=h^{2}+k^{2}\)

\(x^{2}+y^{2}-2 h x-2 k y=0\)

\(\because\) passes through \((1,0)\)

\(\Rightarrow 1+0-2 \mathrm{h}=0\)

\(\Rightarrow \mathrm{h}=1 / 2\)

\(\because \ \mathrm{OC}=\frac{\mathrm{OP}}{2}\)

\(\sqrt{\left(\frac{1}{2}\right)^{2}+\mathrm{k}^{2}}=\frac{3}{2}\) 

\(\frac{1}{4}+\mathrm{k}^{2}=\frac{9}{4}\)

\(\mathrm{k}^{2}=2\)

\(\mathrm{k}= \pm \sqrt{2}\)

\(\therefore\) Possible coordinate of

\(\mathrm{c}(\mathrm{h}, \mathrm{k})\left(\frac{1}{2}, \sqrt{2}\right)\left(\frac{1}{2},-\sqrt{2}\right)\)

\(4\left(\mathrm{~h}^{2}+\mathrm{k}^{2}\right)=4\left(\frac{1}{4}+2\right)=4\left(\frac{9}{4}\right)=9\) 

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

...