Question

What is the equation of the circle with center (0, 0) that passes through the point (–4, –6)?

Answer 1

Answer:

Option 4: x^2+y^2 = 52

Step-by-step explanation:

Given

Centre at origin

Point on circle = (-4, -6)

The distance between origin and point on circle will be the radius of the circle

So,

$r = \sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}\\r= \sqrt{(-4-0)^2+(-6-0)^2} \\r = \sqrt{(-4)^2+(-6)^2}\\r = \sqrt{16+36} \\r = \sqrt{52}$

As the center is at origin the standard equation will be:

$x^2+y^2 = r^2$

Putting the value of r

$x^2+y^2 = (\sqrt{52})^2\\x^2+y^2 = 52$

Hence, last option i.e. Option 4 is correct ..