Question

A circle is centered at the point (5,-4) and passes through the point (-3,2).

Answer 1

Answer: $(x-5)^2+(y+4)^2=100$

Step-by-step explanation:

Equation of circle having center at (h,k) :

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

Then, the equation of circle having center at (5,-4) and radius r will be:-

$(x-5)^2+(y-(-4))^2=r^2\\\\(x-5)^2+(y+4)^2=r^2$   (1)

Since, circle is passes through (-3,2).

Put x= -3 and y=-2

$(-3-5)^2+(2+4)^2=r^2\\\\\Rightarrow\ 64+36=r^2\\\\\Rightarrow\ r^2=100$

Putting value of $r^2=100$ in (1), we get

$(x-5)^2+(y+4)^2=100$

Hence, the equation of this circle is $(x-5)^2+(y+4)^2=100$

Answer 2

Answer:

$(x-5)^2 + (y+4)^2 = 10^2$

Step-by-step explanation:

We need to find the equation of the circle. First, the formula:

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

Where (h,k) is the center and r is the radius

The center is (5,-4), so we can say:

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

Now, to find the radius, we can use the distance formula to find distance between (5,4) and (-3,2).

The distance formula is  $\sqrt{(y_2-y_1)^2 + (x_2-x_1)^2}$

Where

x_1 = 5

x_2 = -3

y_1 = 4

y_2 = 2

Plugging in, we get:

$\sqrt{(2+4)^2 + (-3-5)^2} \\=\sqrt{6^2 + 8^2}\\ =\sqrt{100} \\=10$

Hence, the radius is 10 and we can write the equation of circle as:

$(x-5)^2 + (y+4)^2 = 10^2$