# Find the equation of circle whose centre is (2, -3) and which passes through the point (3, 4) .

12 views
in Circles
closed
Find the equation of circle whose centre is (2, -3) and which passes through the point (3, 4) .
1. x2+ y2- 4x + 6y + 37 = 0
2. x2- y2- 4x + 6y - 37 = 0
3. x2+ y2 - 4x - 6y - 37 = 0
4. x2 + y2- 4x + 6y - 37 = 0

by (30.0k points)
selected

Correct Answer - Option 4 : x2 + y2- 4x + 6y - 37 = 0

Concept:

If Circle with centre (h, k) and passes through an arbitrary point P (x, y) on the circle, then the radius of circle,

| CP | = r $\rm \sqrt{\left ( x-h \right )^{2}+\left ( y-k \right )^{2}}$

Equation of circle having centre (h, k) and radius r is

(x – h) 2 + (y – k) 2 = r

Calculation:

Let C (2, -3) be the centre of of the given circle and let it passe through the point P (3, 4). Then, the radius of circle

| CP | = r = $\rm \sqrt{\left ( 3-2 \right )^{2}+\left ( 4+3 \right )^{2}}$ = $\sqrt{50}$

∴ Required equation of circle is ,

( x - 2 )2 + ( y + 3 )2 =  ( $\sqrt{50}$ )2

x+ y- 4x + 6y - 37 = 0

The correct option is 4.