Question

NO LINKS!! Please help me with this problem 1dd​

Answer 1

`(x+2)^2+(y+2)^2=61`

Explanation:

`\boxed{\begin{minipage}{7.4 cm}\underline{Midpoint between two points}\\\\Midpoint $=\left((x_2+x_1)/(2),(y_2+y_1)/(2)\right)$\\\\\\where $(x_1,y_1)$ and $(x_2,y_2)$ are the endpoints.\\\end{minipage}}`

Given endpoints of the diameter.

• (3, 4)
• (-7, -8)

The center of a circle is the midpoint of its diameter.

Therefore, substitute the given endpoints into the midpoint formula to find the center of the circle:

`\implies \left ((-7+3)/(2),(-8+4)/(2)\right)`

`\implies \left ((-4)/(2),(-4)/(2)\right)`

`\implies \left (-2,-2\right)`

`\boxed{\begin{minipage}{7.4 cm}\underline{Distance between two points}\\\\$d=√((x_2-x_1)^2+(y_2-y_1)^2)$\\\\\\where $(x_1,y_1)$ and $(x_2,y_2)$ are the two points.\\\end{minipage}}`

The radius of a circle is the distance between the center and one of the endpoints of its diameter.

Substitute the found center and one of the given endpoints into the distance formula to calculate the radius of the circle:

`\implies r=√((-2-3)^2+(-2-4)^2)`

`\implies r=√((-5)^2+(-6)^2)`

`\implies r=√(25+36)`

`\implies r=√(61)`

`\boxed{\begin{minipage}{5cm}\underline{Equation of a circle}\\\\$(x-a)^2+(y-b)^2=r^2$\\\\where:\\ \phantom{ww}$\bullet$ $(a, b)$ is the center. \\ \phantom{ww}$\bullet$ $r$ is the radius.\\\end{minipage}}`

Substitute the found center and radius into the equation of a circle formula:

`\implies (x-(-2))^2+(y-(-2))^2=\left(√(61)\right)^2`

`\implies (x+2)^2+(y+2)^2=61`

Therefore the equation of the circle in standard form with the given characteristics is:

`\boxed{(x+2)^2+(y+2)^2=61}`