Question

Write the equation of a circle with center (6,4) that passes through the coordinate (2,1). In your final answer, include all of your calculations.

Answer 1

We are given with a circle of center at (6,4) and that passes through (2,1). In this case, we follow the formula (x-h)^2 + (y-k)^2 = r^2.Substituting, (x-6)^2 + (y-4)^2 = r^2. at (2,1), we substitute the points: r then is equal to 5 which is the radius of the circle. The final equation of the circle is then  (x-6)^2 + (y-4)^2  = 25. 

Answer 2

Answer:

$(x-6)^2+(y-4)^2=5^2$

Step-by-step explanation:

We know that the equation of a circle is $(x-h)^2+(y-k)^2=r^2$, with center (h,k).

In this case we have that the center of the circle is (6,4), then replacing that in the equation of the circle:

$(x-6)^2+(y-4)^2=r^2$

Now we need to find the radius of the circle, we have the point (2,1) that passes through the circle and to find the radius we are going to replace the point in the last equation:

$(x-6)^2+(y-4)^2=r^2\\(2-6)^2+(1-4)^2=r^2\\(-4)^2+(-3)^2=r^2\\16+9=r^2\\25=r^2\\\sqrt{25}=r\\5=r$

Then the equation of a circle with center (6,4) is $(x-6)^2+(y-4)^2=5^2$