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  • <img alt="\mathrm{Let\; x^{2}+y^{2}+Ax+By+C= 0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BLet%5C%3B%20x%5E%7B2%7D&plus;y%5E%7B2%7D&plus;Ax&plus;By&plus;C%3D%

\mathrm{Let\; x^{2}+y^{2}+Ax+By+C= 0} be a circle passing through \mathrm{\left ( 0,6 \right )} and touching the parabola \mathrm{y= x^{2}\, \, at\left ( 2,4 \right )}. Then \mathrm{A+C} is equal to ______.

Option: 1

16


Option: 2

88/5


Option: 3

72


Option: 4

-8


Answers (1)

best_answer

\mathrm{0+36+6B+C= 0}
\mathrm{6B+C= -36 \quad \cdots (i)}

\mathrm{4+16+2A+4B+C= 0}
\mathrm{2A+4B+C= -20 \quad \cdots (ii)}

Also,
\mathrm{2x+2y\, {y}'+A+B{y}'= 0}
\mathrm{{y}')_{\left ( 2,4 \right )}= \frac{-A-4}{8+B}= \left ( 2x \right )_{x= 2}}
\mathrm{\frac{-\left ( A+4 \right )}{B+8}= 4}
\mathrm{A+4B= -36 \quad \cdots (iii)}

\mathrm{A= \frac{-4}{5},\; B= \frac{-44}{5},\; \; C= \frac{84}{5}}

\mathrm{\therefore \; \; A+C= 16}

The correct answer is option (A)

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