To find the point in the first quadrant where the line [tex]\( y = 1.5x + 2 \)[/tex] intersects the circle with radius 6 and center [tex]\( (0,2) \)[/tex], we need to solve the intersection of these two equations.
1. Equation of the line: [tex]\( y = 1.5x + 2 \)[/tex]
2. Equation of the circle: [tex]\( (x - 0)^2 + (y - 2)^2 = 6^2 \)[/tex]
Let's follow these steps to find the intersection point:
### Step 1: Substitute the Line's Equation into the Circle's Equation
Since the equation of the line gives us [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex], we substitute [tex]\( y \)[/tex] from the line equation into the circle equation.
[tex]\[ (x - 0)^2 + ( (1.5x + 2) - 2 )^2 = 6^2 \][/tex]
### Step 2: Simplify the Equation
Simplify the equation inside the circle formula:
[tex]\[ x^2 + (1.5x)^2 = 36 \][/tex]
### Step 3: Expand and Combine Like Terms
Expand the squared term and simplify:
[tex]\[ x^2 + 2.25x^2 = 36 \][/tex]
[tex]\[ 3.25x^2 = 36 \][/tex]
### Step 4: Solve for [tex]\( x \)[/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[ x^2 = \frac{36}{3.25} \][/tex]
[tex]\[ x^2 \approx 11.076923 \][/tex]
[tex]\[ x \approx \sqrt{11.076923} \][/tex]
[tex]\[ x \approx 3.328201177351375 \][/tex]
### Step 5: Find [tex]\( y \)[/tex]
Using [tex]\( x \approx 3.328201177351375 \)[/tex], substitute back into the line equation to find [tex]\( y \)[/tex]:
[tex]\[ y = 1.5x + 2 \][/tex]
[tex]\[ y = 1.5 \times 3.328201177351375 + 2 \][/tex]
[tex]\[ y \approx 6.9923017660270625 \][/tex]
### Final Answer
The point of intersection in the first quadrant is approximately:
[tex]\[
\begin{array}{l}
x = 3.328201177351375 \\
y = 6.9923017660270625
\end{array}
\][/tex]
