Find the equation of a circle with PQ as diameter, where P(2,-6) and Q(-6,-4)

y=-¼x+⁴⁴/8

Step-by-step explanation:

p(2,-6). Q(-6-4)

p (x¹,y¹). Q(x²,y²)

1) find the gradient(slope)

gradient=y²-y¹

x²-x¹

gradient = -4-(-6)

-6-2

gradient= -4+6

-6-2

gradient=²/-8

2)equation is supposed to be in the form of y=mx+c

where m is the gradient

thus.

gradient=y²-y¹

x²-x¹

when finding the equation we only use one of the values of the coordinates leaving the other as an unknown value.

2 = y-(-4)

-8 x-(-6)

2 =y+4

-8. x+6

remove the denominators by cross multiplying that is:

2(x+6)= -8(y+4)

2x+12= -8y-32

express in the format y=mx+c

2x+12+32=-8y

2x+44=-8y.

divide all sides by -8 to remain with y

²/-8x+⁴⁴/-8=y

-¼x+ -⁴⁴/8=y

y=-¼x-⁴⁴/8

c.

step-by-step explanation:

she should of added 9^2 and 12^2. that is because it is a^2+b^2=c^2. c^2 is the hypotenuse and should always be bigger than the other two numbers. which is why she should of added 9 squared and 12 squared together instead of 15 squared.