Question

Find the equation of the circle which passes through the points (3,-2) (-2,0)and and having its centre on the line 2x-y-3=0

Answer 1

Answer:

(x + 1.5)² + (y + 6)² = 36.25

Step-by-step explanation:

$1^{st}$ - Slope of the line passing through A(3, - 2) and B(- 2, 0) is

$m_{AB}$ = $\frac{-2-0}{3+2}$ = $-\frac{2}{5}$

Equation of the line passing through points A and B is

y = $-\frac{2}{5}$ ( x + 2) ⇔ y = $-\frac{2}{5}$ x  $-\frac{4}{5}$ ......... (1)

$2^{nd}$ - Find equation of the perpendicular bisector to segment AB.

m = $-\frac{1}{m_{AB} }$ = $\frac{5}{2}$

Coordinates of midpoint of segment AB are

( $\frac{x_{A} +x_{B} }{2}$ , $\frac{y_{A} +y_{B} }{2}$ ) = ( $\frac{1}{2}$ , - 1 )

y + 1 = $\frac{5}{2}$ ( x - $\frac{1}{2}$ ) ⇔ y = $\frac{5}{2}$ x - $\frac{9}{4}$ ......... (2)

$3^{rd}$ - Find the coordinates of intersection of 2x - y - 3 = 0 and (2)

2x - y - 3 = 0 ⇔ y = 2x - 3

y = $\frac{5}{2}$ x - $\frac{9}{4}$  

2x - 3 =  $\frac{5}{2}$ x - $\frac{9}{4}$  ⇔ 8x - 12 = 10x - 9 ⇒ x = - 1.5

y = 2(- 1.5) - 3 = - 6

Thus, the center of the circle is O( - 1.5, - 6)

$4^{th}$ - Find OB = OA = r

d = $\sqrt{(x_{A} -x_{O})^2 +(y_{A} -y_{O} )^2 }$ = $\sqrt{(-2+1.5)^2 +(0 +6)^2}$ = $\sqrt{36.25}$

The last step:  The standard equation of a circle is (x - h)² + (y - k)² = r² ,

(h, k) are coordinates of a center of the circle.

(x + 1.5)² + (y + 6)² = 36.25