Question

Write the standard form of the equation of the circle with center (7,1) that also contains the point (−1,−5).

Answer 1

Answer:

$(x-7)^2+(y-1)^2=100$

Step-by-step explanation:

1) Find the radius

We can do this by using the distance equation with the centre (7,1) and the given point (-1,-5):

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ where the two points are $(x_1,y_1)$ and $(x_2,y_2)$

Plug in the points (7,1) and (-1,-5)

$d=\sqrt{(-1-7)^2+(-5-1)^2}\\d=\sqrt{(-8)^2+(-6)^2}\\d=\sqrt{64+36}\\d=\sqrt{100}\\d=10$

Therefore, the radius of the circle is 10 units.

2) Plug the data into the equation of a circle

Equation of a circle (when not centred at the origin):

$(x-h)^2+(y-k)^2=r^2$ where the centre is $(h,k)$ and r is the radius

Plug in the centre (7,1) as (h,k)

$(x-7)^2+(y-1)^2=r^2$

Plug in the radius 10

$(x-7)^2+(y-1)^2=10^2\\(x-7)^2+(y-1)^2=100$

I hope this helps!