Answer:
[tex] \rm\displaystyle (x + 2{)}^{2} + (y + 5 {)}^{2} = {1}^{} [/tex]
[tex] \rm\displaystyle (x - 7{)}^{2} + (y + 3{)}^{2} = { 7 }[/tex]
Step-by-step explanation:
QUESTION-1:
we are given the center and the redious of a circle equation
remember that,
[tex] \rm\displaystyle E _{c} :(x - h {)}^{2} + (y - k {)}^{2} = {r}^{2} [/tex]
where (h,k) is the centre coordinate and r is redious of the circle
given that, h=-2 , k=-5 and r=1
Thus substitute:
[tex] \rm\displaystyle (x - ( - 2){)}^{2} + (y - ( - 5) {)}^{2} = {1}^{2} [/tex]
simplify:
[tex] \rm\displaystyle (x + 2{)}^{2} + (y + 5 {)}^{2} = {1}^{} [/tex]
QUESTION-2:
Likewise
[tex] \rm\displaystyle E _{c} :(x - h {)}^{2} + (y - k {)}^{2} = {r}^{2} [/tex]
given that,h=7,k=-3 and r=√7
substitute:
[tex] \rm\displaystyle (x - (7{))}^{2} + (y - ( - 3){)}^{2} = { \sqrt{7} ^{2} }[/tex]
simplify:
[tex] \rm\displaystyle (x - 7{)}^{2} + (y + 3{)}^{2} = { 7 }[/tex]
