Answer:
A. [tex](x+5)^2+(y-4)^2=16[/tex]
B. [tex](x-1)^2+(y-4)^2=2\\[/tex]
C. [tex]$\left(x-3\right)^2+\left(y-5\right)^2=9$[/tex]
Step-by-Step Explanation:
A. Tangent to x-axis, r=4, contains (-5,8)
The equation of a circle is this: [tex](x-h)^2+(y-k)^2=r^2\\[/tex]
Therefore, we know that our equation will be: [tex](x-h)^2+(y-k)^2=4^2=16[/tex]
Because the circle is tangent to the x-axis, the point (-5,8) will be the top of the circle.
Therefore, the center is (-5,4), where -5=h and 4=k.
The answer is [tex](x+5)^2+(y-4)^2=16[/tex]
B. Diameter endpoints (2,5) and (0,3)
To solve for the diameter, we can use this equation: [tex]d=\sqrt{(x_{1}-x_{2})^2+(y_{1}-y_{2})^2 }[/tex]
[tex]d=\sqrt{(2-0)^2+(5-3)^2 }\\\\d=\sqrt{2^2+2^2}=\sqrt{4+4} =\sqrt{8}= 2\sqrt{2}[/tex]
Because the radius is half of the diameter, [tex]r=\sqrt{2}[/tex]
Therefore, we know that our equation will be: [tex](x-h)^2+(y-k)^2=(\sqrt{2}) ^2=2[/tex]
Then, we can decipher that h=1, as the midpoint of 0 and 2 is 1, and k=4, as the midpoint of 3 and 5 is 4.
The answer is [tex](x-1)^2+(y-4)^2=2[/tex]
C. Center on x=3, tangent to y-axis at (0,5)
We can see that the center is (3,5). Therefore, the equation is [tex](x-3)^2+(y-5)^2=r^2\\[/tex]
To solve for r, we can substitute the point (0,5) into the equation.
[tex](0-3)^2+(5-5)^2=r^2\\(-3)^2+(0)^2=r^2\\r^2=9[/tex]
The answer is [tex](x-3)^2+(y-5)^2=9[/tex]
Hope this helped!
