Test Index

Circles Part 2

Section: Mathematics

© examsnet.com

Question : 84 of 87

Marks: +1, -0

Let

C

be a circle passing through the points

A(2,-1)

B(3,4)

. The line segment

A B

s not a diameter of

C

r

is the radius of

C

and its centre lies on the circle

(x-5)^2+(y-1)^2=\frac{13}{2}

r^2

[26-Jun-2022-Shift-1]

32
$\frac{65}{2}$
$\frac{61}{2}$
30

Solution: 👈: Video Solution

Equation of perpendicular bisector of

A B

is

y-\frac{3}{2}=-\frac{1}{5}\left(x-\frac{5}{2}\right) \Rightarrow x+5y=10

Solving it with equation of given circle,

(x-5)^2\left(\frac{10-x}{5}-1\right)^2=\frac{13}{2}

\Rightarrow (x-5)^2\left(1+\frac{1}{25}\right)=\frac{13}{2}

\Rightarrow x-5=\pm \frac{5}{2} \Rightarrow x=\frac{5}{2} \text{ or } \frac{15}{2}

But

x \neq \frac{5}{2}

because

AB

is not the diameter.

So, centre will be

\left(\frac{15}{2}, \frac{1}{2}\right)

Now,

r^2=\left(\frac{15}{2}-2\right)^2+\left(\frac{1}{2}+1\right)^2

=\frac{65}{2}

© examsnet.com

Go to Question:

Prev Question

Next Question