Answer:
[tex](x-4)^2+(y-5)^2=49[/tex]
Step-by-step explanation:
So we want to find the equation of a circle with the center at (4,5) and which passes through the point (-3,5).
First, recall the standard form for a circle, given by the equation:
[tex](x-h)^2+(y-k)^2=r^2[/tex]
Where (h,k) is the center and r is the radius.
We already know that the center is (4,5). So, substitute 4 for h and 5 for k. Therefore, our equation is now:
[tex](x-4)^2+(y-5)^2=r^2[/tex]
Now, we need to find the radius.
Remember that our circle passes through the point (-3,5). In other words, when x is -3, y is 5. So, substitute -3 for x and 5 for y to solve for r. Therefore:
[tex](-3-4)^2+(5-5)^2=r^2[/tex]
Subtract within the parentheses:
[tex](-7)^2+(0)^2=r^2[/tex]
Square:
[tex]49=r^2[/tex]
Square root:
[tex]r=7[/tex]
Therefore, the radius is 7.
So, substitute 7 into our equation, we will acquire:
[tex](x-4)^2+(y-5)^2=(7)^2[/tex]
Square:
[tex](x-4)^2+(y-5)^2=49[/tex]
So, our equation is:
[tex](x-4)^2+(y-5)^2=49[/tex]
And we're done!
