Answer:
The point of intersection is:
[tex]\displaystyle \left(\frac{6\sqrt{5}}{5}, \frac{3\sqrt{5}}{5}+5\right)\approx \left(2.68, 6.34)[/tex]
Step-by-step explanation:
We want to find the point in QI at which the line with the equation:
[tex]y=0.5x+5[/tex]
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:
[tex](x-h)^2+(y-k)^2=r^2[/tex]
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:
[tex](x-0)^2+(y-5)^2=(3)^2[/tex]
Simplify:
[tex]x^2+(y-5)^2=9[/tex]
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:
[tex]x^2+((0.5x+5)-5)^2=9[/tex]
Simplify:
[tex]x^2+(0.5x)^2=9[/tex]
Square:
[tex]x^2+0.25x^2=9[/tex]
Combine like terms:
[tex]\displaystyle 1.25x^2=\frac{5}{4}x^2=9[/tex]
Solve for x:
[tex]\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}[/tex]
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:
[tex]\displaystyle y=0.5\left(\frac{6\sqrt{5}}{5}\right)+5=\frac{3\sqrt{5}}{5}+5\approx 6.34[/tex]
So, the point of intersection in QI is:
[tex]\displaystyle \left(\frac{6\sqrt{5}}{5}, \frac{3\sqrt{5}}{5}+5\right)\approx \left(2.68, 6.34)[/tex]
