Answer:
[tex]\left(x\right)^2+\left(y\right)^2=10[/tex]
Step-by-step explanation:
Given that center of the circle is at (0,0) and it passes through a point (-1,-3).
Now we need to write the equation of the circle with given center and point on the circle.
So let's plug given values into standard formula of the circle
[tex]\left(x-h\right)^2+\left(y-k\right)^2=r^2[/tex]
there (h,k) gives center of the circle then h=0, k=0
So we get equation :
[tex]\left(x-0\right)^2+\left(y-0\right)^2=r^2[/tex]
[tex]\left(x\right)^2+\left(y\right)^2=r^2[/tex] ...(i)
plug the given point (-1,-3) that is x=-1 and y=-3 into (i)
[tex]\left(-1\right)^2+\left(-3\right)^2=r^2[/tex]
[tex]1+9=r^2[/tex]
[tex]10=r^2[/tex]
plug above value into (i)
[tex]\left(x\right)^2+\left(y\right)^2=10[/tex]
Hence final answer is [tex]\left(x\right)^2+\left(y\right)^2=10[/tex]
