Question

Find the equation of the circle having (-5, 6) and (3,-4) the end points of a diameter.​

Answer 1

First we must find the center of the circle, which is the midpoint of the diameter.

The midpoint of a line segment whose endpoints are (x1,y1) and (x2,,y2) is

M = ((x1 + x2))/2,(y1 + y2)/2).

So, the midpoint of the segment whose endpoints are (-5,-4) and (3,-6) is ((-5 + 3)/2,(-4 + -6)/2) = (-1,-5)

The point (-1,-5) is the center of the circle.

Next, we must find the length of the radius of the circle, which is the segment starting at the center and extending to a point on the circle. The center is (-1,-5), and we know that (-5,-4) and (3,-6) are points on the circle. I will show the calculation of the radius using each point on the circle, even though you only need to use one point.

If we use (-5,-4), the distance formula tells us that the distance from (-1,-5) to (-5,-4) is

√(-5 - (-1))2 + (-4 - (-5))2 = √16 + 1 = √17

So of the radius of the circle is √17.

The equation for a circle with center at (a,b) and radius r is

(x - a)2 + (y - b)2 = r2

Now we will substitute a = -1, b = -5, and r = √17

The equation of the circle is (x - (-1))2 + (y - (-5))2 = (√17)2.

Simplifying, we (x + 1)2 + (y + 5)2 = 17

Step-by-step explanation:

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Answer 2

Answer:

The equation of the circle having (-5, 6) and (3,-4) as the end points of a diameter is x²+y²+2x-2y-39 = 0

Step-by-step explanation:

Given,

The endpoints of a diameter of a circle are (-5, 6) and (3,-4)

To find,

The equation of the circle

Recall the formula

The equation of the circle is  $(x -h)^2 + (y -k)^2 = r^2$ -----------------(1)

, where (h,k) is the center of the circle and 'r' is the radius of the circle

The distance between two points $A(x_1,y_1) \ and \ B(x_2,y_2)$ is given by

AB = $\sqrt{(x2- x1)^2 + (y2 -y1)^2}$ ----------------------(2)

The midpoint of the line joining  $A(x_1,y_1) \ and \ B(x_2,y_2)$ is given by

$(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}) }$ ----------------------(3)

Solution:

Since the points (-5, 6) and (3,-4) are the endpoints of a diameter,

The diameter is the distance between the two points (-5, 6) and (3,-4)

Using the formula (2),

Distance between the points (-5, 6) and (3,-4) = $\sqrt{(3- (-5))^2 + ((-4) -6)^2}$

= $\sqrt{8^2 + (-10)^2}$

= $\sqrt{64+ 100}$

= $\sqrt{164}$

= $\sqrt{2^2 X 41}$

= 2√41

∴The diameter of the circle = 2√41

∴The radius of the circle =√41

Since the points (-5, 6) and (3,-4) are the endpoints of a diameter of the circle, then the center of the circle is the mid point of the line joining  (-5, 6) and (3,-4)

Using formula (3) we get

Center of the circle = $(\frac{-5+3}{2}, \frac{6 +(-4)}{2}) }$

The center of the circle is (-1,1)

∴The equation of the circle with center (-1,1) and radius √41 is given by

Substitute  h = -1 and k = 1 and r  = √41 in equation (1) we get

(x-(-1))² + (y-1)² = (√41)²

(x+1)² + (y-1)² = 41

Expanding we get

x²+2x+1+y²-2y+1 = 41

x²+y²+2x-2y-39 = 0

∴  The equation of the circle having (-5, 6) and (3,-4) the end points of a diameter is x²+y²+2x-2y-39 = 0

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