Question

Find the equation of a circle with a point at ( 10 , - 4 ) and a point at ( -2 , - 4 )​

Answer 1

`\rm\displaystyle (x - 4) ^(2) + {(y + 4)}^(2) = 36`

Explanation:

the given points are the diameter points of circle because notice that in the both points y coordinate is the same therefore it's a horizontal diameter

since (10,-4),(-2,-4) are the diameter points of the circle the midpoint of the diameter will be the centre of the circle

remember midpoint formula,

`\displaystyle M = \left( (x _(1) + x_(2) )/(2) , ( y_(2) + y_(2))/(2) \right)`

let,

• 
  `\displaystyle x _(1) = 10`
• 
  `\displaystyle x _(2) = - 2`
• 
  `\displaystyle y _(1) = - 4`
• 
  `\displaystyle y _(2) = -4`

thus substitute:

`\rm\displaystyle M = \left( (10 + ( - 2))/(2) , ( - 4 + ( - 4))/(2) \right)`

simplify addition:

`\rm\displaystyle M = \left( (8)/(2) , ( - 8)/(2) \right)`

simplify division:

`\rm\displaystyle M = \left( 4, - 4 \right)`

so the centre of the circle is (4,-4)

since it's a horizontal diameter the the redious will be the difference between the x coordinate of the Midpoint and the any x coordinate of the given two points but I'll use (-2,-4) therefore the redious is

`\displaystyle r = 4 - ( - 2)`

simplify which yields:

`\displaystyle\boxed{ r =6}`

recall the equation of circle

`\displaystyle (x - h) ^(2) + {(y - k)}^(2) = {r}^(2)`

we acquire that,

• h=4
• k=-4
• r=6

therefore substitute:

`\rm\displaystyle (x - 4) ^(2) + {(y - ( - 4))}^(2) = {6}^(2)`

simplify:

`\rm\displaystyle (x - 4) ^(2) + {(y + 4)}^(2) = 36`

and we are done!

also refer the attachment

(the graph is web resource of desmos)

Answer 2

Solution given:

letA=(10,-4)

B=(-2,-4)

centre[C](h,k)=
`(10-2)/(2),(-4-4)/(2)=(+4,-4)`

radius=
`√((4-10)²+(-4+4)²)=6`units

we have

Equation of a circle is;

(x-h)²+(y-k)²=r²

(x-4)²+(y+4)²=36

or.

x²-8x+16+y²+8y+16=36

x²-8x+8y+y²=36-32

x²-8x+8y+y²=4

The equation is (x-4)²+(y+4)²=36 or x²-8x+8y+y²=4.