Test Index

Circles Part 1

Section: Mathematics

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Question : 22 of 100

Marks: +1, -0

The line

2x-y+1=0

is a tangent to the circle at the point

(2,5)

and the centre of the circle lies on

x-2y=4

. Then, the radius of the circle is

[17 Mar 2021 Shift 1]

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

Solution:

Given,

2x-y+1=0

is a tangent to the circle at

(2,5)

So, normal at

(2,5)

will be

\;\frac{y-5}{x-2} = \;\frac{-1}{2}

⇒

2y-10 \;\; = -x+2

x+2y \;\; =12

Now, it is also given that centre lies on

x-2y=4

.

So, coordinates of centre will be the solution of

\begin{cases} x+2y=12 \\ x-2y=4 \end{cases}

\Rightarrow x=8, y=2

Radius will be the distance between

(8,2)

and

(2,5)

r^{2}=(8-2)^{2}+(2-5)^{2}

\Rightarrow\; r^{2}\; = 36+9

\Rightarrow\; r\; = \sqrt{45}=3\sqrt{5}

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