Question

Write the equation of a circle with endpoints of the diameter at (-1,-6) and (-7,4)?

Answer 1

`x^2+y^2+8x+2y-17=0`

Explanation:

we need the coordinates of the centre and the length of the radius
some step are omitted for the reader to check for themselves
centre

The centre coordinates will be the midpoints of the end points of the diameter

centre`=>((-1+ -7)/2, (-6 +4)/2)`

`=(-8/2, -2/2)=(-4,-1)`

radius

this is half the length of the diameter

by Pythagoras

radius`=1/2sqrt((-1 - -7)^2+(-6 -4)^2`

`r=1/2sqrt(6^2+(-10)^2)`

`r=1/2sqrt(136)=sqrt34`

the eqn of a circle with centre`(a,b)` and radius `r`

`(x-a)^2+(y-b)^2=r^2`

`(x- -4)^2+(y- -1)^2=34`

`(x+4)^2+(y+1)^2=34`

`x^2+8x+16+y^2+2y+1=34`

`x^2+y^2+8x+2y-17=0`