Question

Aakarshini Nemali You wrote 19 hours ago. ​hi mrs. e, i was wondering if you could help me understand this question from the usa test prep, If (5, 4) and (7, 2) are the endpoints of a diameter of a circle, what is the equation of the circle? A) (x − 6)2 + (y + 3)2 = 2 B) (x − 6)2 + (y − 3)2 = 2 C) (x + 6)2 + (y +3)2 = 8 D) (x + 6)2 + (y − 3)2 = 8

Answer 1

Answer:

$x^{2} +y^{2}-6y-12x+43=0$

Step-by-step explanation:

The general form of the equation of a circle is given as:

$(x-a)^2+(y-b)^2=r^2$ where (a,b) are the centre and r is the radius.

If (5, 4) and (7, 2) are the endpoints of a diameter, we find the coordinate of the centre using the midpoint formula.

$\frac{1}{2}(x_1+x_2 , y_1+y_2) \:where (x_1,y_1 \:and\: x_2, y_2)$ are the two points.

Midpoint of (5, 4) and (7, 2) = $\frac{1}{2}(5+7 , 4+2) =\frac{1}{2}(12 , 6)=(6,3)$

Next, we determine the radius. Using the distance formula, we find the distance from the centre to any of the endpoints.

Distance from (5,4) to (6,3)

r=$\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

=$\sqrt{(5-6)^2+(4-3)^2} \\=\sqrt{(-1)^2+(1)^2}=\sqrt{1+1}=\sqrt{2}\\r=\sqrt{2}$

The equation of the circle centre (6,3) with radius $\sqrt{2}$ is therefore given as:

$(x-a)^2+(y-b)^2=r^2\\(x-6)^2+(y-3)^2=(\sqrt{2}) ^2\\x^{2} +y^{2}-6y-12x+45=2\\x^{2} +y^{2}-6y-12x+45-2=0\\x^{2} +y^{2}-6y-12x+43=0$