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  • Find the centre and radius of the smaller of the two circles that touch the parabola <img alt="\mathrm{75 y^2=64(5 x-3) \: at \: \left(\frac{6}{5}, \frac{8}{5}\right)}" src="https://entrancecorner.

Find the centre and radius of the smaller of the two circles that touch the parabola \mathrm{75 y^2=64(5 x-3) \: at \: \left(\frac{6}{5}, \frac{8}{5}\right)}and the x-axis.
 

Option: 1

(1,2) ; 2
 


Option: 2

(1,2) ; 1
 


Option: 3

(2,1) ; 2
 


Option: 4

(2,1) ; 1


Answers (1)

best_answer

The parabola can be written as

\mathrm{ 75 y^2=32(10 x-6) }

\mathrm{ Tangent\: at \: \left(x_1, y_1\right)\: is \: 75 y y_1=32\left[5\left(x+x_1\right)-6\right] }

\mathrm{ 75 y \cdot \frac{8}{5}=32\left[5\left(x+\frac{6}{5}\right)-6\right] at \left(\frac{6}{5}, \frac{8}{5}\right) }

\mathrm{ or \: 4 x-3 y=0 }

Above is common tangent to parabola and circle at \mathrm{ \left(\frac{6}{5}, \frac{8}{5}\right) }.The centre of the circle will be on normal at a distance \mathrm{ \pm r } from the point of contact. Slope of tangent is \mathrm{ \frac{4}{3} } and hence of normal will be \mathrm{ \frac{-3}{4}=\tan \theta }

\mathrm{ \therefore \cos \theta=-\frac{4}{5}, \sin \theta=\frac{3}{5} }

Hence the co-ordinates of centres are points \mathrm{ C_1\: and \: C_2 } on the normal through \mathrm{ P\left(\frac{6}{5}, \frac{8}{5}\right) } and at a distance \mathrm{ r \: \: or -r \: \: from \: \: P. }

\mathrm{ \begin{aligned} & \frac{x-\frac{6}{5}}{-\frac{4}{5}}=\frac{y-\frac{8}{5}}{\frac{3}{5}}=\underset{C_1}{r} C_2 \\ \therefore \quad & C_1=\left(\frac{-4 r+6}{5}, \frac{3 r+8}{5}\right) \end{aligned} }

\mathrm{ C_2=\left(\frac{4 r+6}{5}, \frac{-3 r+8}{5}\right) }
Since the circle touches x-axis therefore \mathrm{ y } co-ordinate of centre is radius
\mathrm{ \begin{aligned} & \therefore \quad \frac{3 r+8}{5}=r . \quad \therefore \quad r=4 \text { for } C_1 \text { (Bigger) } \\ & \text { or } \frac{-3 r+8}{5}=r . \quad \therefore \quad r=1 \text { for } C_2 \text { (Smaller) } \end{aligned} }

\mathrm{ Hence\: bigger\: =(-2,4), 4 \: and \: smaller \: =(2,1), 1. }

Hence option 4 is correct.
 

Posted by

Nehul

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