Respuesta :
Answer:
[tex]y=-4x+12\\y =-0.421x+12[/tex]
Step-by-step explanation:
The lines through (0,12) would have equation of the form
y = mx+12
If this is tangent, the distance of centre from the tangent = radius of the circle
Distance of (6,5) ie centre from tangent line is
[tex]\frac{6m+7}{\sqrt{1+m^2} } =\sqrt{17} \\(6m+7)^2 = 17(1+m^2)\\[/tex]
[tex]36m^2+84m+49 =17+17m^2\\19m^2+84m+32=0[/tex]
m=-4,-0.421
Hence tangents are
[tex]y=-4x+12\\y =-0.421x+12[/tex]
The equation of the tangents can be found from the equation of the
perpendicular distance of a point to a line.
Responses:
- The equation of L1 is y = 12 - 4·x
- [tex]The \ equation \ of \ L2 \ is \ \underline{y =12 -\dfrac{8}{19} \cdot x}[/tex]
How can the equations of the tangent lines to a circle be found?
The given equation is; (x - 6)² + (y - 5)² = 17
The tangents of the circle are; L1 and L2
The point of intersection of the lines = (0, 12)
Required;
The equations of L1 and L2
Solution;
The general form of the equation of a circle is; (x - h)² + (y - k)² = r²
Which gives;
The center of the circle, (h, k) = (6, 5)
The radius of the circle, r = √(17)
The y-intercept of the tangent line (the point where x = 0) is given as the point (0, 12), which gives;
The y-intercept, c = 12
The equation of the tangent (straight) lines, L1 and L2 are therefore;
y = m₁·x + 12
y = m₂·x + 12
Where;
m₁, and m₂, are the slopes of the tangent lines
The perpendicular distance from a point (x₁, y₁) to a line a·x + b·y + c = 0 is given as follows;
- [tex]d = \mathbf{\pm \left(\dfrac{a \cdot x_1 + b \cdot y_1 + c}{\sqrt{a^2 + b^2} } \right)}[/tex]
The equation of the tangent are; y = m·x + 12
Therefore;
m·x - y + 12 = 0
The perpendicular distance from the center to the line is given as follows where;
d = The perpendicular distance from the center to the tangents = The radius, r = √(17)
a = m
b = -1
c = 12
Which gives;
[tex]\sqrt{17} = \pm \left(\dfrac{m \cdot 6 - 5 + 12}{\sqrt{m^2 + (-1)^2} } \right) = \mathbf{\pm \left(\dfrac{m \cdot 6+7}{\sqrt{m^2 + 1} } \right)}[/tex]
[tex]\sqrt{17} \times \sqrt{m^2 + 1} = m \cdot 6+7[/tex]
17·(m² + 1) = (m·6 + 7)²
17·m² + 17 = 36·m² + 84·m + 49
36·m² + 84·m + 49 - (17·m² + 17) = 0
19·m² + 84·m + 32 = 0
By using a graphing calculator, we get;
[tex]\mathbf{(m + 4) \cdot \left(m + \dfrac{8}{19} \right) }= 0[/tex]
Therefore;
The slopes of the lines are;
m₁ = -4, and m₂ = [tex]-\dfrac{8}{19}[/tex]
The equation of the tangent lines, y = m·x + 12, are therefore;
L1; y = -4·x + 12 = 12 - 4·x
- L1 is y = 12 - 4·x
[tex]L2 \ is \ y = -\dfrac{8}{19} \cdot x + 12 = \mathbf{12 -\dfrac{8}{19} \cdot x}[/tex]
- [tex]L2 \ is \ \underline{y =12 -\dfrac{8}{19} \cdot x}[/tex]
Learn more about the distance from a point given the coordinates here:
https://brainly.com/question/387459
