Respuesta :
Answer:
The equation of the circle with center (0, 1) and a radius of 10 is [tex]\bold{x^{2} + y^{2} - 2y-99 = 0}[/tex]
Solution:
Equation of circle whose center (h, k) and radius r is given as,
[tex](x-h)^{2} + (y-k)^{2} = r^{2}[/tex] ---- eqn 1
From question, given that the circle has radius 10 and center (0, 1). We have to find the equation of the circle.
Hence we get, h = 0 ; k = 1; r = 10
By using eqn 1, the equation of circle whose center (0, 1) and radius 10 is given as
[tex](x-0)^{2} + (y - 1)^{2} = 10^{2}[/tex]
On Simplifying we get
[tex]x^{2} + (y-1)^{2} = 10^{2}[/tex]
Expand [tex](y-1)^{2}[/tex] using the formula [tex](a-b)^{2} = a^{2} - 2ab + b^{2}[/tex]
[tex]x^{2} + y^{2} - 2y + 1 = 10^{2}[/tex]
[tex]x^{2} + y^{2} - 2y + 1 = 100[/tex]
[tex]x^{2} + y^{2} - 2y + 1-100 = 0[/tex]
[tex]x^{2} + y^{2}- 2y- 99 = 0[/tex]
Hence equation of the circle with center (0, 1) and a radius of 10 is [tex]\bold{x^{2}+y^{2}-2 y-99=0}[/tex]
