Question

Find an equation of the circle that has center (-6, 6) and passes through (3,-4)​

Answer 1

(x+6)² + (y-6)² = 181

Explanation:

equation of the circle in general is (x - a)² + (y - b)² = r²

center is (a, b) in our task a=-6 b=6

so equation of the circle is (x+6)² + (y-6)² = r²

we have point (3,-4)​ ⇒ x=3 y=-4

put x y in equation of the circle ⇒ (3+6)² + (-4-6)² = r² ⇒r² =181

⇒The equation of a circle is (x+6)² + (y-6)² = 181

Answer 2

x²+ y² - 109 = 0

Explanation:

The coordinate of centre of circle = (-6 , 6)

Point through which it passes = (3 , -4)

So , Firstly find the distance between the centre and that point that will be equal to the radius . Using Distance Formula we have ,

⇒ D = √ [ ( x - x' )² + ( y + y')² ]

⇒ D = √ [ (3+6)² + (6+4)² ]

⇒ D = √ [ 9² + 10² ]

⇒ D = √ [ 81 + 100 ]

⇒ D = √ 181

⇒ D = 13.45 .

Now substituting the respective values in the general equation of circle .

⇒ ( x - h)² + (y - k)² = r²

• Where (h,k) is the centre and r is the radius.

⇒ { x - (-6) }² + { y - (6) }² = (13.45)²

⇒ ( x + 6 )² + ( y - 6)² = 13.45²

⇒ x² + 36 + 12x + y² + 36 -12x = 181

⇒ x² + y² + 72 = 181

⇒ x² + y² + 72 - 181 = 0

⇒ x² + y² - 109 = 0

Therefore the equation of the circle is x² + y² - 109 = 0 .