Question

Recall the equation for a circle with center (h,k) and radius r. At what point in the first quadrant does not line with equation y=2x+4 intersects the circle with radius 4 and center (0,4)

Answer 1

(1.79, 7.58)

Explanation:

Standard form equation of a circle with center (h,k) and radius r is

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

Use h = 0, k = 4 and r=4 to give

-->
`\displaystyle{(x-0)^2+(y-4)^2=4^2}`

-->
`x^2 + (y-4)^2 = 16`

`(y-4)^2 = y^2 -8y + 16`

The line is
`y = 2x + 4`

Substitute for this value of y in Equation (1)

`x^2 + (2x + 4 - 4)^2 = 16`

`x^2 + (2x)^2 = 16`

`x^2 + 4x^2 = 16`

`5x^2 = 16`

`x^2 = (16)/(5)`

`x = \pm \sqrt{(16)/(5)}`

`x = \pm (4)/(√(5))`

Since we are asked to find point of intersection only on the first quadrant, we ignore the negative value of x

So
`x = (4)/(√(5) ) = 1.78885 = 1.79` (rounded to 2 decimal places)

Substituting this value of x in
`y = 2x + 4`

`y = 2(1.79) = 4 = 7.58`

So the intersection point is at

(1.79, 7.58)

See attached graph