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

Answer:

(1.79, 7.58)

Step-by-step 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)

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

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

$x^2 + 4x^2 = 16$

$5x^2 = 16$

$x^2 = \frac{16}{5}$

$x = \pm \sqrt{\frac{16}{5}}$

$x = \pm \frac{4}{\sqrt{5}}$

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

So $x = \frac{4}{\sqrt{5} } = 1.78885 = 1.79$  (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