Question

Find the equation for the circle with center (4,5) and passing through (-3,5).

Answer 1

Answer:

$(x-4)^2+(y-5)^2=49$

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:

$(x-h)^2+(y-k)^2=r^2$

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:

$(x-4)^2+(y-5)^2=r^2$

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:

$(-3-4)^2+(5-5)^2=r^2$

Subtract within the parentheses:

$(-7)^2+(0)^2=r^2$

Square:

$49=r^2$

Square root:

$r=7$

Therefore, the radius is 7.

So, substitute 7 into our equation, we will acquire:

$(x-4)^2+(y-5)^2=(7)^2$

Square:

$(x-4)^2+(y-5)^2=49$

So, our equation is:

$(x-4)^2+(y-5)^2=49$

And we're done!