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KEAM 2012 Math Paper

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Question : 82 of 120

Marks: +1, -0

If (-2, 2) and (k, 0) are two diametrically opposite points of a circle of radius 1, then the equation of the circle is

$x^2+y^2+2x-4y+4=0$
$x^2+y^2+4x-2y-4=0$
$x^2+y^2-4x+2y+4=0$
$x^2+y^2+4x-2y+4=0$
$x^2+y^2-4x-2y-4=0$

Solution:

since,

P(-2,2)

and

Q(k,0)

is the end points of a diameter.

\therefore

Mid-point

M

of

PQ=\frac{-2+k}{2},\;\frac{2+0}{2}

i.e.,

M=\left(\frac{k}{2}-1,1\right)

Also,

MP=

radius of circle

\therefore \sqrt{\left(-2-\frac{k}{2}+1\right)^2+(2-1)^2}=1

\Rightarrow \sqrt{\left(-1-\frac{k}{2}\right)^2+1}=1

\Rightarrow 1+\frac{k^2}{4}+k+1=1

\Rightarrow k^2+4k+4=0

\Rightarrow (k+2)^2=0

\Rightarrow

k=-2

Mid-point

M=\left(-\frac{2}{2}-1,1\right)=(-2,1)

which is equal to the centre of circle.

\therefore

Equation of circle is

(x+2)^2+(y-1)^2=1^2

\Rightarrow x^2+4x+4+y^2-2y+1=1

\Rightarrow x^2+y^2+4x-2y+4=0

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