Question

How do you write an equation for a circle with diameter has endpoints(-3,4) and (4,-3)?

Answer 1

The equation of the circle is:

`color(brown)(=>49/2=(x_i-1/2)^2+(y_i-1/2)^2)`

Explanation:

The centre will be at the middle of these points so it is the mean value.

Let point 1 be `P_1->(x_1,y_1)->(-3,4)`
Let point 2 be `P_2->(x_2,y_2)->(4,-3)`
Let the centre be `P_c->(x_x,y_c)`
Let the radius be `r`

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine the centre")`

Mean for x`-> (4+(-3))/2= 1/2 ->x_c`

Mean for y`->(4+(-3))/2=1/2->y_c`

`color(blue)("Centre "-> (x_c,y_c)->(1/2,1/2))`

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine the magnitude of the radius")`
Using Pythagoras

I chose `(x_1,y_1)` to determine the length of `r`

`r^2=(x_c-x_1)^2+(y_c-y_1)^2`

`r^2=[1/2-(-3)]^2+[1/2-4]^2`

`r=sqrt(49/4+49/4) = sqrt(49/2)`

`r=7/sqrt(2)`

Multiply by 1 but in the form of `1=sqrt(2)/sqrt(2)`

`color(blue)(r=(7sqrt(2))/2" "->" " r^2=49/2)`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Putting it all together")`

Let any `x` be `x_i`
Let any `y` be `y_i`

The equation of this circle is

`r^2=(x_i-x_x)^2+(y_i-y_c)^2`

`color(blue)(=>49/2=(x_i-1/2)^2+(y_i-1/2)^2)`