Question

Someone pls help me do these

Answer 1

`\textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \hspace{5em}\stackrel{center}{(\underset{1}{h}~~,~~\underset{-4}{k})}\qquad \stackrel{radius}{\underset{2}{r}} \\\\[-0.35em] ~\dotfill\\\\ ( ~~ x - 1 ~~ )^2 ~~ + ~~ ( ~~ y-(-4) ~~ )^2~~ = ~~2^2\implies {\large \begin{array}{llll} (x-1)^2+(y+4)^2=4 \end{array}}`

now as for the 2nd one, we know the center and a point on it, well, what's the radius? well, recall that for a circle the radius is simply the distance from the center to the circle in proper, so the distance between those two points is the radius

`~~~~~~~~~~~~\textit{distance between 2 points} \\\\ C(\stackrel{x_1}{-2}~,~\stackrel{y_1}{5})\qquad P(\stackrel{x_2}{-2}~,~\stackrel{y_2}{-1})\qquad \qquad d = √(( x_2- x_1)^2 + ( y_2- y_1)^2) \\\\\\ CP=√((~~-2 - (-2)~~)^2 + (~~-1 - 5~~)^2) \implies CP=√((-2 +2)^2 + (-1 -5)^2) \\\\\\ CP=√(( 0 )^2 + ( -6 )^2) \implies CP=√( 0 + 36 ) \implies CP=√( 36 )\implies \stackrel{radius}{CP=6}`

so we're really looking for the equation of a circle with a center at (-2 , 5) with a radius of 6

`\textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \hspace{5em}\stackrel{center}{(\underset{-2}{h}~~,~~\underset{5}{k})}\qquad \stackrel{radius}{\underset{6}{r}} \\\\[-0.35em] ~\dotfill\\\\ ( ~~ x - (-2) ~~ )^2 ~~ + ~~ ( ~~ y-5 ~~ )^2~~ = ~~6^2\implies {\large \begin{array}{llll} (x+2)^2+(y-5)^2=36 \end{array}}`

Answer 2

`1.\ (x - 1)^2 + (y + 4)^2 = 4\\2.\ (x + 2)^2 + (y -5)^2 = 36`