Question

What is the equation of the circle with center (3, 5) that passes through the point (–4, 10)?

Answer 1

Answer:

Option 1:

$(x-3)^2+(y-5)^2 = 74$

Step-by-step explanation:

Given

Centre: (h,k) = (3,5)

Point on circle = (-4,10)

The distance between Centre and point on circle will be the radius of the circle

The distance formula will be used to calculate distance

$r = \sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}\\r = \sqrt{(-4-3)^{2}+(10-5)^{2}}\\r=\sqrt{(-7)^{2}+(5)^{2}}\\r=\sqrt{49+25}\\r=\sqrt{74}$

The standard equation of circle is:

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

Putting the values of h,k and r

$(x-3)^2+(y-5)^2 = (\sqrt{74})^2\\(x-3)^2+(y-5)^2 = 74$

Hence, first option is correct ..