Question

What is the equation of the circle when given the endpoints of a diameter (-2, 15) and (6, -13)?

Answer 1

Answer:

$(x-2)^2+(y-1)^2=212$

or

$x^2 - 4 x + y^2 - 2 y - 207 = 0$

Step-by-step explanation:

Method 1:

For finding the radius:

The distance between points $(x_{1} ,y_{1})$and $(x_{2} ,y_{2})$ is:

$\sqrt{(x_{1} -y_{1})^{2}+(x_{2} -y_{2})^{2}}$

Substitute $(x_{1} ,y_{1})$ = (-2, 15) and $(x_{2} ,y_{2})$ = (6, -13):

$=\sqrt{((-2-6)^2+(-(-13)+15)^2}$

 $=4\sqrt{53}$

Which is the diameter

The radius:

$4\sqrt{53} /2\\=2\sqrt{53}$

The centre:

The midpoint of two endpoints:

$x=(-2+6)/2=2\\y=(15-13)/2=1$

Center is (2,1)

The standard form:

Circle with radius r and centre (a,b) is $(x-a)^2 + (y-b)^2 = r^2$

$(x-2)^2+(y-1)^2=(2\sqrt{53})^2\\ (x-2)^2+(y-1)^2=212$

Expanding to get the general form:

$x^2 - 4 x + y^2 - 2 y - 207 = 0$

Method 2:

Let the locus $P(x,y)$ be the circle:

By angle in a semi-circle, The angle inscribed in a semicircle is always a right angle (90°).

This means line joining from one endpoint of diameter is perpendicular to the line joining from another endpoint, two lines always have the product of slope = -1:

$(\frac{y-15}{x+2} )(\frac{y+13}{x-6})=-1\\ (y-15)(y+13)=-(x+2)(x-6)\\x^2+y^2-4x-2y-207=0$