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

`x^(2) +y^(2)-6y-12x+43=0`

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.

`(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) =
`(1)/(2)(5+7 , 4+2) =(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=
`√((x_1-x_2)^2+(y_1-y_2)^2) \\`

=
`√((5-6)^2+(4-3)^2) \\=√((-1)^2+(1)^2)=√(1+1)=√(2)\\r=√(2)`

The equation of the circle centre (6,3) with radius
`√(2)` is therefore given as:

`(x-a)^2+(y-b)^2=r^2\\(x-6)^2+(y-3)^2=(√(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`