Question

Sat question: a circle in the xy-plane has a diameter with endpoints (-1,-3) and (7,3). if the point (0,b) lies on the circle and b > 0, what is the value of b

Answer 1

Answer:

The value of b is:

                          b = 4

Step-by-step explanation:

We are given the endpoints of a diameter as:

 (-1,-3) and (7,3)

Now, we know that the center of the circle lies in between the endpoints of the diameter of the circle.

Let the coordinates of center be: (x,y)

We know that if a point (x,y) lie in between (a,b) and (c,d) then the coordinates of point (x,y) is given by:

$x=\dfrac{a+c}{2},\ y=\dfrac{b+d}{2}$

Hence, the center of circle has coordinates:

$x=\dfrac{-1+7}{2},\ y=\dfrac{-3+3}{2}\\\\x=\dfrac{6}{2},\ y=\dfrac{0}{2}\\\\x=3,\ y=0$

Hence, the center of circle is located at  (3,0).

Also, the length of diameter with the help of distance formula is:

$Diameter=\sqrt{(7-(-1))^2+(3-(-3))^2}\\\\Diameter=\sqrt{8^2+6^2}\\\\Diameter=\sqrt{64+36}\\\\Diameter=\sqrt{100}\\\\Diameter=10\ units$

Now, we know that the length of radius is half the length of diameter.

Hence, Radius= 5 units

Hence, the equation of circle is:

$(x-3)^2+(y-0)^2=5^2$

i.e.

$(x-3)^2+y^2=25$

( Since, the equation of the circle with center (h,k) and radius r is given by:

$(x-h)^2+(y-k)^2=r^2$   )

Now, the point (0,b) lie on the circle i.e. the point satisfies the equation of circle.

Hence, we put (0,b) in the circle.

$(0-3)^2+b^2=25\\\\9+b^2=25\\\\b^2=25-9\\\\b^2=16\\\\b=\pm 4$

But b>0

Hence, we get:

$b=4$

Answer 2

1) use the distance formula to find that the radius (r) of the circle = 5

2) use the midpoint formula to find that the center of the circle (h,k) = (3, 0)

3) Now use the formula of a circle and input the (h,k) and r to create:

(x - 3)² + (y - 0)² = 5² → (x - 3)² + y² = 25

4) input the "x" value given in the question (x = 0) and solve for "y":

(0 - 3)² + y² = 25 → 9 + y² = 25 → y² = 16 → y = +/- 4

Since the question states that "b" must be positive, you can disregard the -4.

Answer: b = 4