Knowing the extreme points of a diameter in circle are #A(-3,1) and D(2,5)#, find the canonical equation of a circle?

Answer 1

#(x+1/2)^2+(y-3)^2=41/4#

The equation of a circle with center #(x_0,y_0)# and radius #r# is
#(x-x_0)^2+(y-y_0)^2=r^2#
To find this equation in the given case, then, we need the radius and the center. As we are given the endpoint of a diameter, and the radius of a circle is half of the length of a diameter, we can find the radius as half of the distance from #A# to #D#.
The distances between two points #(x_1,y_1)# and #(x_2,y_2)# is given by #sqrt((x_2-x_1)^2+(y_2-y_1)^2)#. So we have
#r = 1/2"dist"(A,D) = sqrt((2-(-3))^2+(5-1)^2)/2 = sqrt(41)/2#
Next, we find the center of the circle by using the fact that the midpoint of any diameter of the circle will be its center point. The midpoint of the line segment connecting #(x_1,y_1)# and #(x_2,y_2)# is #((x_1+x_2)/2,(y_1+y_2)/2)#. So we have the center as
#((-3+2)/2,(1+5)/2) = (-1/2,3)#

We can enter our values to find the circle's equation now that we know the radius and center.

#(x-(-1/2))^2 + (y-3)^2 = (sqrt(41)/2)^2#

or

#(x+1/2)^2+(y-3)^2=41/4#
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Answer 2

To find the canonical equation of a circle, we need the coordinates of its center and its radius.

  1. Center of the Circle: The center of the circle lies at the midpoint of the diameter. To find the midpoint, we can use the midpoint formula: [ \text{Midpoint} = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) ] Let's denote the coordinates of points A and D as ( (x_1, y_1) ) and ( (x_2, y_2) ) respectively.

  2. Radius of the Circle: The radius can be found by calculating the distance between the center of the circle and any of the points on the circumference.

Given the points A(-3,1) and D(2,5), let's find the center and the radius:

  1. Center of the Circle: Using the midpoint formula: [ \text{Midpoint} = \left( \frac{-3 + 2}{2}, \frac{1 + 5}{2} \right) ] [ \text{Midpoint} = \left( \frac{-1}{2}, 3 \right) ]

  2. Radius of the Circle: We can choose any of the points, say A(-3,1), and calculate the distance to the center: [ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ] [ \text{Distance} = \sqrt{(-\frac{1}{2} - (-3))^2 + (3 - 1)^2} ] [ \text{Distance} = \sqrt{\frac{25}{4} + 4} ] [ \text{Distance} = \sqrt{\frac{41}{4}} ]

So, the center of the circle is ( \left( -\frac{1}{2}, 3 \right) ) and the radius is ( \sqrt{\frac{41}{4}} ).

The canonical equation of a circle is: [ (x - h)^2 + (y - k)^2 = r^2 ] where ( (h, k) ) are the coordinates of the center, and ( r ) is the radius.

Substitute the values we found: [ \left( x + \frac{1}{2} \right)^2 + \left( y - 3 \right)^2 = \left( \sqrt{\frac{41}{4}} \right)^2 ]

This is the canonical equation of the circle.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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