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  • The centre of the circle passing through the point and touching the curve <img alt="\mathrm{y=x^{2}\: at \: (2,

The centre of the circle passing through the point (0,1) and touching the curve \mathrm{y=x^{2}\: at \: (2,4)} is
 

Option: 1

\left(-\frac{16}{5}, \frac{27}{10}\right)

 


Option: 2

\left(-\frac{16}{7}, \frac{53}{10}\right)
 


Option: 3

\left(-\frac{16}{5}, \frac{53}{10}\right)
 


Option: 4

none of these


Answers (1)

best_answer

Let the centre be \mathrm{(h, k), \: then\: (h-0)^2+(k-1)^2=(h-2)^2+(k-4)^2}

\mathrm{ \Rightarrow \quad 4 h+6 k=19 }...........(i)
Again, centre of the circle must lie on the normal to the parabola \mathrm{ y=x^2 \: at \: (2,4) }

\Rightarrow equation of normal is \mathrm{y-4=-\frac{1}{4}(x-2)}

\mathrm{ \Rightarrow \quad h+4 k=18 }       ..........(ii)

From (i) and (ii) \mathrm{ \Rightarrow h=-\frac{16}{5}, k=\frac{53}{10}}

Hence option 3 is correct.
 

Posted by

Pankaj Sanodiya

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