Answer:
[tex](x+8)^2+(y-8)^2=64[/tex]
Step-by-step explanation:
Required conditions:
- Tangent to both axes.
- Center in the second quadrant.
- Radius = 8 units.
If the circle is tangent to both axes, its center will be the same distance from both axes. That distance is its radius.
If the center of the circle is in quadrant II, the center will have a negative x-value and a positive y-value → (-x, y).
Therefore, the coordinates of the center will be (0-r, 0+r) where r is the radius.
If the radius is 8 units, then the center is (-8, 8).
[tex]\boxed{\begin{minipage}{4 cm}\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}}[/tex]
Substitute the found center and given radius into the formula to create an equation for the circle that satisfies the given conditions:
[tex]\implies (x-(-8))^2+(y-8)^2=8^2[/tex]
[tex]\implies (x+8)^2+(y-8)^2=64[/tex]
