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.
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]
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]
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]
Answer:
Step-by-step explanation:
For a, we need to find the slope of the given line, because the slope for the line PN has a perpendicular slope. We are given the line y = 1/3x; therefore, the slope of that line is 1/3 and the perpendicular slope is -3. Using that slope and the coordinates of the center (7,4) we can find the equation of line PN:
y-4=-3(x-7) so
y-4=-3x+21 and
y=-3x+25
b is a little trickier. If we want to know the equation for circle C we need the length of the radius. We will use a system of equations to find the length of that radius by using a system of equations to find the coordinates of P. The 2 lines we have that point P go through are
[tex]y=\frac{1}{3}x+0[/tex] and [tex]y=-3x+25[/tex]
We could use substitution and set them equal to each other and solve for x:
[tex]\frac{1}{3}x=-3x+25[/tex] Multiply everything by 3 to get rid of the fraction and get
x = -9x + 75 and
10x = 75 so
x = 7.5
Plug this back in to solve for y:
[tex]y=\frac{1}{3}*7.5[/tex]
y = 2.5
Those are the coordinates for point P (7.5, 2.5).
Now we can find the length of the radius by using the standard form of a circle and the coordinates of the center along with those for point P:
[tex](7.5-7)^2+(2.5-4)^2=r^2[/tex] and
[tex](.5)^2+(-1.5)^2=r^2[/tex] and
[tex].25+2.25=r^2[/tex] so
[tex]r^2=2.5[/tex]
The equation for the circle is
[tex](x-7)^2+(y-4)^2=2.5[/tex]