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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

Sagot :

Nayefx

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]

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