What is the equation of a circle with diameter AB; A is (5,4) and B is (-1,-4)?

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First calculate the radius of the circle:

The horizontal distance between the points is the distance between the x coordinates of A and B, |5| + |-1| = 6, and the vertical distance is the distance between the y coordinates of A and B, |4| + |-4| = 8

The diameter is then the direct distance between A and B which is sqrt(6^2 + 8^2) = sqrt(36+64) = sqrt(100) = 10. This makes the radius 5.

We also need to know where the center of the circle is. This has coordinates halfway between the coordinates of A and B. Half the horizontal distance between A and B is 3 and half the vertical distance is 4, so that the center is at C(x,y) = B(x,y) + (3,4) = (-1,-4) + (3,4) = (2,0) (adding (3,4) again gives the coordinates of B).

The equation for a circle written in standard form is given by

`(x-x_c)^2 + (y-y_c)^2 = r^2`

where `r` is the radius and `(x_c,y_c)` are the coordinates of the center C.

Therefore the circle whose diameter is AB can be written in standard form as

`(x-2)^2 + y^2 = 25`

**answer**

Since we have learned in geometry that angle in semi circle is right angle. Apply this result and can calculate equation of circle.

Let P(x,y) be any point on the circle.Thus A(5,4),B(-1,-4) and P(x,y) on the circle.

Also PA is perpendicular to PB.

slope of PA=(y-4)/(x-5)

slope of PB=(y+4)/(x+1)

PA is perpendicular to PB ,therefore

`((y-4)/(x-5))((y+4)/(x+1))=-1`

`(y-4)(y+4)=-(x-5)(x+1)`

`(x-5)(x+1)+(y-4)(y+4)=0`

`x^2-4x-5+y^2-16=0`

`x^2+y^2-4x-21=0`

which is required circle.

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