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Circle

Section: Mathematics

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Question : 29 of 89

Marks: +1, -0

A circle passing through the point

(1,0)

makes an intercept of length 4 units on X -axis and an intercept of length

2 \sqrt{11}

units on Y - axis. If the centre of the circle lies in the fourth quadrant, then the radius of the circle is

[AP EAPCET 22 May 2025 S2]

$4 \sqrt{5}$
$3$
$2 \sqrt{5}$
$5$

Solution:

Let the

x

-intercepts be

(a, 0)

and

(a+4,0)

Let the

y

-intercepts be (

0, b

) and

(0, b+2 \sqrt{11})

Center of circle

=(a+2, b+\sqrt{11})

Since, the center is in the fourth quadrant

h>0

and

k<0

\Rightarrow a+2>0 \;\text{ and }\; b+\sqrt{11}<0

Equation of circle

(x-h)^2+(y-k)^2=r^2

\;\Rightarrow (1-(a+2))^2+(0-(b+\sqrt{11}))^2=r^2

\;\Rightarrow (-a-1)^2+(-b-\sqrt{11})^2=r^2

Substituting

(a, 0)

gives

\; (a-(a+2))^2+(0-(b+\sqrt{11}))^2=r^2

\;\Rightarrow (-2)^2+(-b-\sqrt{11})^2=r^2

substituting (

a+4,0

) gives

\;\Rightarrow (a+4-(a+2))^2+(0-(b+\sqrt{11}))^2=r^2

\;\Rightarrow 2^2+(-b-\sqrt{11})^2=r^2

Similarly,

(0, b)

gives

(0-(a+2))^2+(b-(b+\sqrt{11}))^2=r^2

\;\text{ substituting }\;(0, b+2 \sqrt{11})

(0-(a+2))^2+(b+2 \sqrt{11}-(b+\sqrt{11}))^2=r^2

(-a-2)^2+(\sqrt{11})^2=r^2

Solving above equation, we get

\; r^2=20

\; r=2 \sqrt{5}

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