Question

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

Answer 1

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 = r2   

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 

⇒ x² + y² - 4x + 6y - 37 = 0. 

The correct option is 4.