Question

Find the equation for the circle with a diameter whose endpoints are (3,-1) and (-2,-1)

Answer 1

Answer:

The equation for the circle is: $(x-\frac{1}{2} )^2 + y^2 = (\frac{5}{2} )^2$

Step-by-step explanation:

Here, let us assume the two end points of the diameter are:

P (3,-1) and Q (-2,-1)

Now, as we knew diameter is the chord that passer through the center of the circle.

⇒ Mid point of DIAMETER  = Center coordinates  of the Circle

Let us assume O(x,y) is the center of the line  segment PQ.

So, by MID POINT FORMULA:

$(x,y)= (\frac{3 -2}{2} , \frac{-1-1}{2} ) =( \frac{1}{2},\frac{-2}{2}) \\\implies (x,y) =( 0.5 , -1)$

⇒ The center coordinates of the circle  = O(0.5,-1)   ...... (1)

Now, RADIUS = Half of DIAMETER

Using DISTANCE FORMULA:

$PQ = \sqrt{(3-(-2))^2 + (-1 - (-1))^2} = \sqrt{(5)^2 + 0} = 5$

So, Radius  = 5/2  = 2.5 units

Now, the  equation of circle with radius r and center coordinate ( h,k) is given as: $(x-h)^2 + (y-k)^2 = r^2$

Substitute r = 2.5 and (h,k)  = (0.5,0) we get:

$(x-0.5)^2 + (y-0)^2 = (2.5)^2\\\implies (x-\frac{1}{2} )^2 + y^2 = (\frac{5}{2} )^2$

Hence the equation for the circle is: $(x-\frac{1}{2} )^2 + y^2 = (\frac{5}{2} )^2$