Answer:
Option 4: x^2+y^2 = 52
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= √((-4-0)^2+(-6-0)^2) \\r = √((-4)^2+(-6)^2)\\r = √(16+36) \\r = √(52)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/ffemqwww2nsfey9e1x5ilupix2h0t6ib62.png)
As the center is at origin the standard equation will be:
![x^2+y^2 = r^2\\](https://img.qammunity.org/2020/formulas/mathematics/middle-school/n1p9m03f3675pikui9tpg04e22rnt0lp3w.png)
Putting the value of r
![x^2+y^2 = (√(52))^2\\x^2+y^2 = 52](https://img.qammunity.org/2020/formulas/mathematics/middle-school/nrb084xy73pkxuscm9xptnjwzre2dsiadn.png)
Hence, last option i.e. Option 4 is correct ..