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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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Find the point of intersection of the line $4x-3y-10=b$ and circle $x^2+y^2-2x+4y-20=0$

$\begin{array}{1 1}(A)\;(4,2)\;and\;(-2,-6)\\(B)\;(4,0)\;and\;(-2,-6)\\(C)\;(4,2)\;and\;(2,6)\\(D)\;(4,-2)\;and\;(-2,6)\end{array}$

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1 Answer

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$4x-3y-10=0$ -----(1)
$x^2+y^2-2x+4y=20$------(2)
Solving (1) and (2) we get,
$\bigg(\large\frac{3y+10}{4}\bigg)^2+y^2-2 \bigg( \large\frac{3y+10}{4}\bigg)$$+4y-20=0$
=> $9y^2+60\;y+100 +16y^2-24y-80+64\;y-320=0$
=>$25y^2+100y-300=0$
=> $y^2+4y-12=0$
=> $y=2,-6$
=> $x=4,-2$
Points are $(4,2)$ an $(-2,-6)$
Hence A is the correct answer.
answered Apr 8, 2014 by meena.p
 

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