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KEAM 2013 Math Paper

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Question : 60 of 120

Marks: +1, -0

The equation of the circle which touches the lines

x = 0, y = 0,

4x + 3y = 12

is

$x^{2}+y^{2}-2x-2y-1=0$
$x^{2}+y^{2}-2x-2y+3=0$
$x^{2}+y^{2}-2x-2y+2=0$
$x^{2}+y^{2}-2x-2y+1=0$
$x^{2}+y^{2}-2x-2y-3=0$

Solution:

Let the radius of the circle be

r

.

since, circle touches the lines

x=0, y=0,

i.e.,

x

-axis and

y

-axis, then its centre is

C \equiv (r, r)

Now, radius

(P C)=

Perpendicular distance from (C) to the line

4 x+3 y=12

i.e.,

\;\; r = \frac{\left| 4r + 3r - 12 \right|}{\sqrt{16+9}}

\Rightarrow \;\; 7r - 12 = \pm 5r

So

r = 1

or

6

But

r \neq 6

\therefore

Required equation of circle is

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

\Rightarrow \;\; x^{2}+1-2x+y^{2}+1-2y=1

\Rightarrow \;\; x^{2}+y^{2}-2x-2y+1=0

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