Question

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

Answer 1

Answer:

Solution given:

letA=(10,-4)

B=(-2,-4)

centre[C](h,k)=$\frac{10-2}{2},\frac{-4-4}{2}=(+4,-4)$

radius=$\sqrt{(4-10)²+(-4+4)²}=6$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.

Answer 2

Answer:

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

Step-by-step 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,

$\displaystyle M = \left( \frac{x _{1} + x_{2} }{2} , \frac{ y_{2} + y_{2}}{2} \right)$

let,

• $\displaystyle x _{1} = 10$
• $\displaystyle x _{2} = - 2$
• $\displaystyle y _{1} = - 4$
• $\displaystyle y _{2} = -4$

thus substitute:

$\rm\displaystyle M = \left( \frac{10 + ( - 2)}{2} , \frac{ - 4 + ( - 4)}{2} \right)$

simplify addition:

$\rm\displaystyle M = \left( \frac{8}{2} , \frac{ - 8}{2} \right)$

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)