Question

Recall the equation for a circle with center (h,k) and radius r . At what point in the first quadrant does the line with equation y=1.5x+5 intersect the circle with radius 4 and center (0, 5)?

Answer 1

The form of the equation of the circle is

$(x-h)^2+(y-k)^2=r^2$

(h, k) is the center

r is the radius

Since the given center is (0, 5) and the given radius is 4, then

The equation of the circle is

$\begin{gathered} (x-0)^2+(y-5)^2=16 \\ x^2+(y-5)^2=16\rightarrow(1) \end{gathered}$

Since the equation of the line is

$y=1.5x+5\rightarrow(2)$

Substitute y in equation (1) by equation (2)

$\begin{gathered} x^2+(1.5x+5-5)^2=16 \\ x^2+(1.5x)^2=16 \\ x^2+2.25x^2=16 \end{gathered}$

Add the like terms on the left side

$3.25x^2=16$

Divide both sides by 3.25

$\begin{gathered} \frac{3.25x^2}{3.25}=\frac{16}{3.25} \\ x^2=\frac{64}{13} \end{gathered}$

Take a square root for both sides

$\begin{gathered} \sqrt{x^2}=\sqrt{\frac{64}{13}} \\ x=2.218800785 \end{gathered}$

Substitute the value of x in equation (2) to find y

$\begin{gathered} y=1.5(2.218800785)+5 \\ y=8.328201177 \end{gathered}$

Round them to 3 decimal places

x = 2.219

y = 8.328