Question

Given the equation of the circle is x²+ y²+ 10x + 2y-10=0 find the following: a. center b. radius c Circumference in terms of T d. the area to the nearest tenth.​

Answer 1

a) center = (-5, -1)

b) radius = 6

c) circumference = 12π

d) area = 113.1 units² (nearest tenth)

Explanation:

`\boxed{\begin{minipage}{6.5cm}\underline{Standard form of an 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}}`

Given equation:

`x^2+ y^2+ 10x + 2y-10=0`

Collect like terms and move the constant to the right side of the equation:

`\implies x^2+10x+y^2+2y=10`

Add the square of half the coefficient of the terms in x and y to both sides of the equation:

`\implies x^2+10x+\left((10)/(2)\right)^2+y^2+2y+\left((2)/(2)\right)^2=10+\left((10)/(2)\right)^2+\left((2)/(2)\right)^2`

Simplify:

`\implies x^2+10x+25+y^2+2y+1=10+25+1`

`\implies x^2+10x+25+y^2+2y+1=36`

Factor the perfect square trinomials in x and y on the left side of the equation:

`\implies (x+5)^2+(y+1)^2=36`

Compare with the standard form of the equation of a circle:

• a = -5
• b = -1
• r² = 36

Therefore:

• center = (a, b) = (-5, -1)
• radius = √(36) = 6

`\boxed{\begin{minipage}{4 cm}\underline{Circumference of a circle}\\\\$C=\pi d$\\\\where:\\ \phantom{ww}$\bullet$ $d$ is the diameter. \\ \end{minipage}}`

The diameter of a circle is twice its radius.

`\begin{aligned}\implies d&=2r\\d&=2 \cdot 6 \\d&= 12\end{aligned}`

To find the circumference of the circle in terms of πd, substitute the found diameter, d = 12, into the circumference formula:

`\implies C=12 \pi`

`\boxed{\begin{minipage}{4 cm}\underline{Area of a circle}\\\\$A=\pi r^2$\\\\where:\\ \phantom{ww}$\bullet$ $r$ is the radius. \\ \end{minipage}}`

To find the area of the circle, substitute the found radius, r = 6, into the area formula:

`\implies A=\pi \cdot 6^2`

`\implies A=36 \pi`

`\implies A=113.097335...`

`\implies A=113.1\;\; \sf units^2\;(nearest\;tenth)`