Question

HELPPPPP ME GUYSSSS........

1. Find the equation of the circle whose center and radius are given.

center ( -2, -5), radius = 1

2. Find the equation of the circle whose center and radius are given.

center ( 7, -3), radius = sqr7

Answer 1

`\rm\displaystyle (x + 2{)}^(2) + (y + 5 {)}^(2) = {1}^{}`

`\rm\displaystyle (x - 7{)}^(2) + (y + 3{)}^(2) = { 7 }`

Explanation:

QUESTION-1:

we are given the center and the redious of a circle equation

remember that,

`\rm\displaystyle E _(c) :(x - h {)}^(2) + (y - k {)}^(2) = {r}^(2)`

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:

`\rm\displaystyle (x - ( - 2){)}^(2) + (y - ( - 5) {)}^(2) = {1}^(2)`

simplify:

`\rm\displaystyle (x + 2{)}^(2) + (y + 5 {)}^(2) = {1}^{}`

QUESTION-2:

Likewise

`\rm\displaystyle E _(c) :(x - h {)}^(2) + (y - k {)}^(2) = {r}^(2)`

given that,h=7,k=-3 and r=√7

substitute:

`\rm\displaystyle (x - (7{))}^(2) + (y - ( - 3){)}^(2) = { √(7) ^(2) }`

simplify:

`\rm\displaystyle (x - 7{)}^(2) + (y + 3{)}^(2) = { 7 }`