JEE MAIN - Mathematics (2004 - No. 43)
If the lines 2x + 3y + 1 + 0 and 3x - y - 4 = 0 lie along diameter of a circle of circumference $$10\,\pi $$, then the equation of the circle is :
$${x^2}\, + \,{y^2} + \,2x\, - \,2y - \,23\,\, = 0$$
$${x^2}\, + \,{y^2} - \,2x\, - \,2y - \,23\,\, = 0$$
$${x^2}\, + \,{y^2} + \,2x\, + \,2y - \,23\,\, = 0$$
$${x^2}\, + \,{y^2} - \,2x\, + \,2y - \,23\,\, = 0$$
Explanation
Two diameters are along
$$2x+3y+1=0$$ and $$3x-y-4=0$$
solving we get center $$(1,-1)$$
circumference $$ = 2\pi r = 10\pi $$
$$\therefore$$ $$r=5$$.
Required circle is, $${\left( {x - 1} \right)^2} + {\left( {y + 1} \right)^2} = {5^2}$$
$$ \Rightarrow {x^2} + {y^2} - 2x + 2y - 23 = 0$$
$$2x+3y+1=0$$ and $$3x-y-4=0$$
solving we get center $$(1,-1)$$
circumference $$ = 2\pi r = 10\pi $$
$$\therefore$$ $$r=5$$.
Required circle is, $${\left( {x - 1} \right)^2} + {\left( {y + 1} \right)^2} = {5^2}$$
$$ \Rightarrow {x^2} + {y^2} - 2x + 2y - 23 = 0$$
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