Test Index

Circles Part 1

Section: Mathematics

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Question : 26 of 100

Marks: +1, -0

If the locus of the mid-point of the line segment from the point

(3,2)

to a point on the circle,

x^{2}+y^{2}=1

is a circle of radius

r

r

[26 Feb 2021 Shift 2]

1
$\frac{1}{2}$
$\frac{1}{3}$
$\frac{1}{4}$

Solution:

Given, circle

x^{2}+y^{2}=1

Let

(h, k)

be the mid-point of line segment from point

(3,2)

to a point on circle. Let point on circle be

(a, b)

. Then,

\;h=\;\frac{a+3}{2} \Rightarrow a=2h-3

\;k=\;\frac{b+2}{2} \Rightarrow b=2k-2

\because (a, b)

is point on circle, then

(a)^{2}+(b)^{2} \;\; =1

\Rightarrow \;\; (2h-3)^{2}+(2k-2)^{2} \;\; =1

\Rightarrow \;\; \left(h-\frac{3}{2}\right)^{2}+(k-1)^{2}=\frac{1}{4} \;\; =\left(\frac{1}{2}\right)^{2}

which is an equation of circle with radius

\frac{1}{2}

.

\therefore

Radius

=\frac{1}{2}

.

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