Answer:
(a) y = -3x +25
(b) (x -7)² +(y -4)² = 2.5
(c) 10/3
Step-by-step explanation:
You want the equation of line PN, the equation of the circle, and the y-intercept of the tangent line parallel to the line shown in the diagram.
(a) Line PN
Line PN will be the line through point N(7, 4) that is perpendicular to the line OP with slope 1/3. PN will have the opposite reciprocal slope, -1/(1/3) = -3, so we can write its equation in point slope form as ...
[tex]y -y_1=m(x -x_1)\\\\y-4=-3(x-7)[/tex]
Rearranging to slope-intercept form, we have ...
[tex]y=-3x+21+4\\\\\boxed{y=-3x+25}[/tex]
(b) Circle equation
To write the circle equation, we need to know the center and radius of the circle. The center is given, and the radius can be found from the formula for the distance from N to P. This makes use of the equation for OP written in general form.
[tex]y=\dfrac{1}{3}x\\\\3y=x\\\\x-3y=0\qquad\text{general form equation for line OP}\\\\|NP|=\dfrac{|x-3y|}{\sqrt{1^2+(-3)^2}}=\dfrac{|7-3(4)|}{\sqrt{10}}=\dfrac{5}{\sqrt{10}}\qquad\text{circle radius}[/tex]
The standard form equation for the circle with center (h, k) and radius r is ...
[tex](x-h)^2+(y-k)^2=r^2\\\\(x-7)^2+(y-4)^2=\left(\dfrac{5}{\sqrt{10}}\right)^2\\\\\\\boxed{(x-7)^2+(y-4)^2=2.5}[/tex]
(c) Parallel tangent
The general form equation for the parallel tangent can be written as ...
x -3y +c = 0
We want the distance of N from this tangent line to be the same as NP, so we have ...
[tex]|NQ|=\dfrac{|x-3y+c|}{\sqrt{10}}=\dfrac{5}{\sqrt{10}}\\\\\\|x-3y+c|=5\\\\|7-3(4)+c|=5\\\\c-5=\pm5\\\\c=5\pm5=\{0,10\}[/tex]
We know line OP has c = 0, so the general form equation for the tangent through point Q is ...
x -3y +10 = 0
Solving for y gives ...
[tex]y=\dfrac{1}{3}x+\dfrac{10}{3}\\\\\\\boxed{k=\dfrac{10}{3}}[/tex]
