[tex]x^2 +(y-3)^2=17[/tex] is the standard form of equation of circle
Solution:
Given that a circle has center (0, 3) and passes through point (-4, 4)
We have to find the standard form of equation of circle
The standard form of equation of circle is given as:
[tex](x-h)^2+(y-k)^2=r^2[/tex] --------- eqn 1
Where the center being at the point (h, k) and the radius being "r"
Here center = (h, k) = (0, 3)
Plug in (h, k) = (0, 3) and (x, y) = (-4, 4) in above equation
[tex](-4-0)^2+(4-3)^2=r^2\\\\16+1 = r^2\\\\r^2 = 17[/tex]
Now plug in [tex]r^2=17[/tex] and (h, k) = (0, 3) in eqn 1
[tex](x-0)^2+(y-3)^2=17\\\\x^2 +(y-3)^2=17[/tex]
Thus the standard form of equation is found
