Question

What is the general form of the equation for the given circle centered at O(0, 0)?

a. x^2 + y^2 + 41 = 0
b. x^2 + y^2 − 41 = 0
c. x^2 + y^2 + x + y − 41 = 0
d. x^2 + y^2 + x − y − 41 = 0

Answer 1

the general form of such an equation is
x^2 + y^2 = r^2 where r = radius

In this case r^2 = 4^2 + 5^2 = 41

So the required equation is x^2 ^ y^2 = 41

B

Answer 2

Option B. x² + y²- 41 = 0

Explanation:

We have to find the equation of a circle with center as (0, 0) and a point on circle is (4, 5).

Since standard equation of a circle is x² + y² = r² with origin(0, 0) as center.

radius r = √(4-0)²+(5-0)² = √(16+25) = √41

Therefore equation of the circle will be

x² + y² = 41

x² + y²- 41 = 0

Option B. is the correct option.