Sure, let’s work through the problem step-by-step:
### Step 1: Identify Point [tex]\( P \)[/tex]
We know that point [tex]\( P \)[/tex] lies on the circle with the equation [tex]\( x^2 + y^2 = 90 \)[/tex], and we are given the [tex]\( x \)[/tex]-coordinate of [tex]\( P \)[/tex] is 3 and the point lies below the [tex]\( x \)[/tex]-axis.
### Step 2: Determine the [tex]\( y \)[/tex]-coordinate of Point [tex]\( P \)[/tex]
To find the [tex]\( y \)[/tex]-coordinate of [tex]\( P \)[/tex], we substitute [tex]\( x = 3 \)[/tex] into the circle's equation:
[tex]\[ x^2 + y^2 = 90 \][/tex]
[tex]\[ 3^2 + y^2 = 90 \][/tex]
[tex]\[ 9 + y^2 = 90 \][/tex]
Subtracting 9 from both sides:
[tex]\[ y^2 = 81 \][/tex]
Taking the square root of both sides and considering the point is below the [tex]\( x \)[/tex]-axis:
[tex]\[ y = -\sqrt{81} \][/tex]
[tex]\[ y = -9 \][/tex]
So, the coordinates of point [tex]\( P \)[/tex] are [tex]\( (3, -9) \)[/tex].
### Step 3: Calculate the Slope of the Radius at [tex]\( P \)[/tex]
The slope of the radius at [tex]\( P \)[/tex] can be obtained using the coordinates of the center of the circle [tex]\((0, 0)\)[/tex] and [tex]\( P(3, -9) \)[/tex]:
[tex]\[ \text{slope of radius} = \frac{\Delta y}{\Delta x} = \frac{-9 - 0}{3 - 0} = \frac{-9}{3} = -3 \][/tex]
### Step 4: Find the Slope of the Tangent Line
The tangent to the circle at [tex]\( P \)[/tex] is perpendicular to the radius at [tex]\( P \)[/tex]. Therefore, the slope of the tangent line is the negative reciprocal of the radius slope:
[tex]\[ \text{slope of tangent line} = -\frac{1}{\text{slope of radius}} = -\frac{1}{-3} = \frac{1}{3} = 0.3333333333333333 \][/tex]
### Step 5: Determine the Equation of the Tangent Line
Using the point-slope form of a line’s equation:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
where [tex]\((x_1, y_1)\)[/tex] is a point on the line [tex]\((3, -9)\)[/tex] and [tex]\(m = 0.3333333333333333\)[/tex] (the slope of the tangent line), we get:
[tex]\[ y - (-9) = 0.3333333333333333(x - 3) \][/tex]
Simplifying:
[tex]\[ y + 9 = 0.3333333333333333x - 1 \][/tex]
Subtract 9 from both sides to get:
[tex]\[ y = 0.3333333333333333x - 10 \][/tex]
### Final Answer:
The equation of the tangent to the circle at the point [tex]\( P \)[/tex] with coordinates [tex]\((3, -9)\)[/tex] is:
[tex]\[ y = 0.3333333333333333x - 10 \][/tex]
or equivalently (for clarity):
[tex]\[ y = \frac{1}{3} x - 10 \][/tex]
