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 = 0.5x + 5 intersect the circle with radius 3 and center (0,5)?
X =
y =​

Answer 1

Answer:

The point of intersection is:

$\displaystyle \left(\frac{6\sqrt{5}}{5}, \frac{3\sqrt{5}}{5}+5\right)\approx \left(2.68, 6.34)$

Step-by-step explanation:

We want to find the point in QI at which the line with the equation:

$y=0.5x+5$

Intersect a circle with a radius of 3 and a center of (0, 5).

First, write the equation of a circle. The equation for a circle is given by:

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

Where (h, k) is the center and r is the radius.

Since our center is (0, 5), h = 0 and k = 5. The radius is 3. So, r = 3. Substitute:

$(x-0)^2+(y-5)^2=(3)^2$

Simplify:

$x^2+(y-5)^2=9$

At the point where the two equations intersect, its x-coordinate and y-coordinate will be the same. Therefore, we can substitute the equation of the line into the equation of the circle and solve for x. So:

$x^2+((0.5x+5)-5)^2=9$

Simplify:

$x^2+(0.5x)^2=9$

Square:

$x^2+0.25x^2=9$

Combine like terms:

$\displaystyle 1.25x^2=\frac{5}{4}x^2=9$

Solve for x:

$\displaystyle \begin{aligned} x^2&=\frac{36}{5} \\ x&=\pm\sqrt{\frac{36}{5}} \\ x&\Rightarrow \frac{6}{\sqrt{5}}=\frac{6\sqrt{5}}{5}\approx2.68\end{aligned}$

Note that since we are looking for the point of intersection in QI, x should be positive. So, we can ignore the negative answer.

To find the y-coordinate, substitute the x-value back into either equation. Using the linear equation:

$\displaystyle y=0.5\left(\frac{6\sqrt{5}}{5}\right)+5=\frac{3\sqrt{5}}{5}+5\approx 6.34$

So, the point of intersection in QI is:

$\displaystyle \left(\frac{6\sqrt{5}}{5}, \frac{3\sqrt{5}}{5}+5\right)\approx \left(2.68, 6.34)$