Question

Find the equation of the circle passing through the three points

(1, 2), (3, - 4) and (5, - 6)

. Also find its centre and radius.

Accepted Answer

Let the required equation of circle be $x^{2} + y^{2} + 2 g x + 2 f y + c = 0 . . . . (1)$

Since the circle $(1)$ passes through the point $(1, 2)$
We have,

$\Rightarrow 2 g + 4 f + c + 5 = 0 . . . . (2)$

circle $(1)$ passes through the point $(3, - 4)$

$\Rightarrow 6 g - 8 f + c + 25 = 0 . . . . . (3)$

Also the point $(5, - 6)$ lies on equation $(1)$ , we get

$\Rightarrow 10 g - 12 f + c + 61 = 0 . . . . (4)$

Equation $(4) -$ Equation $(3)$ , we have

$4 g - 4 f + 36 = 0$

$\Rightarrow g - f + 9 = 0 . . . . . . (5)$

Now, equation $(3) -$ equation $(2)$ , we get

$4 g - 12 f + 20 = 0$

$\Rightarrow g - 3 f + 5 = 0 . . . . . (6)$

Equation $(5) -$ Equation $(6)$ , we get

$2 f + 4 = 0$

$\Rightarrow f = - 2$

Substitute the value of $f$ in equation $(6)$ , we get

$g = - 11$

Now, substitute the value of $f & g$ in equation $(4)$ , we get $10 (- 11) - 12 (- 2) + c + 61 = 0$

$\Rightarrow - 110 + 24 + c + 61 = 0$

$\Rightarrow c = 25$

putting all the values of $g, f and c$ in equation $(1)$

We get, $x^{2} + y^{2} - 22 x - 4 y + 25 = 0$

which is the required equation of circle.

$∴$ Centre of circle be $(- g, - f)$ i.e, $(11, 2)$

and radius of circle $= \sqrt{g^{2} + f^{2} - c}$

$= \sqrt{121 + 4 - 25}$

$= 10 units$

M L Aggarwal Solutions for Chapter: Circles, Exercise 2: EXERCISE

M L Aggarwal Mathematics Solutions for Exercise - M L Aggarwal Solutions for Chapter: Circles, Exercise 2: EXERCISE

Questions from M L Aggarwal Solutions for Chapter: Circles, Exercise 2: EXERCISE with Hints & Solutions

Learn and Prepare for any exam you want !