Question

Find the center and radius of a circle represented by the equation below x²+y²=49

Answer 1

EXPLANATION.

Equation of circle = x² + y² = 49.

As we know that,

General equation of circle,

⇒ x² + y² + 2gx + 2fy + c = 0.

⇒ Centre = (-g,-f).

⇒ Radius = √g² + f² - c = 0.

If the Centre is in origin then the equation of circle is,

⇒ x² + y² = r².

⇒ x² + y² = (7)².

⇒ Centre = (0,0).

⇒ Radius = 7.

                                                                                                                                 

MORE INFORMATION.

Line & circle.

Let L = 0 be a line and S = 0 be a circle, if 'r' be the radius of a circle and p be the length of perpendicular from the Centre of circle on the line, then if.

(1) = p > r ⇒ line is outside the circle.

(2) = p = r ⇒ line touches the circle.

(3) = p < r ⇒ line is the chord of circle.

(4) = p = 0 ⇒ line is diameter of circle.

NOTE :

(1) = Length of the intercept made by the circle on the line is,

2√r² - p².

(2) = The length of intercept made by line y = mx + c with the circle : x² + y² = a² is,

2√a²(1 + m²) -c²/1 + m².

Answer 2

Answer:

Center is (0;0); Radius is 7

Explanation:

Since the equation of the circle also is the following:

$(x - {x _{0} })^{2} + (y - y_{0}) {}^{2} = {r}^{2}$

where r is the radius and$(x _{0}; y_{0} )$

is the center.

Then it is simple to see that the center is (0; 0) and the radius is 7