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  • If the circle <img alt="\mathrm{x^{2}+y^{2}+2 g x+2 f y+c=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bx%5E%7B2%7D&plus;y%5E%7B2%7D&plus;2%20g%20x&plus;2%20f%20y

If the circle \mathrm{x^{2}+y^{2}+2 g x+2 f y+c=0} touches by the line \mathrm{y=x} at point \mathrm{P} such that \mathrm{OP}=6 \sqrt{2}, where \mathrm{O}  is the origin, then the value of \mathrm{c} is equal to

Option: 1

74


Option: 2

62


Option: 3

64


Option: 4

72


Answers (1)

best_answer

The ponit \mathrm{P}  on the circle is \left(0+6 \sqrt{2} \times \frac{1}{2}, 0+6 \sqrt{2} \times \frac{1}{2}\right)

i.e, (6,6) The equation of the circle touching \mathrm{y}=\mathrm{x} at \mathrm{P}(6,6) is 

(\mathrm{x}-6)^{2}+(\mathrm{y}-6)^{2}+\lambda(\mathrm{y}-\mathrm{x})=0
$ $\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{x}(\lambda-12)-\mathrm{y}(\lambda+12)+72=0

This will be same as

\mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{gx}+2 \mathrm{fy}+\mathrm{c}=0
\therefore \quad \lambda-12=2 \mathrm{~g}
-(\lambda+12)=2 \mathrm{f}
\therefore \mathrm{c}=72

Posted by

Pankaj Sanodiya

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