Circle equation and linear equation

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nairb

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Please help solve the question below, thanks!

Q. The coordinates of the centre of the circle C are (6,10). It is given that the x-axis is a tangent to C.
a.) Find the equation of C
b.) The slope and the y-intercept of the straight line L are -1 and k respectively. If L cuts C at A and B, express the coordinates of the mid-point of AB in in terms of k.

Knowing that the answer of A is: x^2+y^2-12x-20y+36=0, please help the b part, thanks.!!
 
What is the equation of the line?
Where does it intersect C, if it does at all?
 
nairb said:
Please help solve the question below, thanks!

Q. The coordinates of the centre of the circle C are (6,10). It is given that the x-axis is a tangent to C.
a.) Find the equation of C
b.) The slope and the y-intercept of the straight line L are -1 and k respectively. If L cuts C at A and B, express the coordinates of the mid-point of AB in in terms of k.

Knowing that the answer of A is: x^2+y^2-12x-20y+36=0, please help the b part, thanks.!!
Click to expand...
For part a.), we know:

"The coordinates of the centre of the circle C are (6,10).
and
It is given that the x-axis is a tangent to C."

What is the length of the radius of the circle described above (C)?

What is the equation of the circle described above (C)?

for part b), what would be the equation of a straight-line whose slope is 'm' and y-intercept is 'b'?
 
The center of the circle is (6, 10) and if it is tangent to the x-axis its radius is 10. The equation is (x- 6)^2+ (y- 10)^2= 100. That does reduce to x^2- 12x+ 36+ y^2- 20y+ 100= 100 or x^2+ y^2- 12x- 20y+ 36= 0 as you say.

Your are told that "the slope and the y-intercept of the straight line L are -1 and k respectively". Do you know what that means? Any (non-vertical) line can be written y= ax+ b. What are a and b?

To determine where the line intersects the circle, replace y in the equation of the circle with "ax+ b" (with the correct values for a and b of course) to get the equation x^2+ (ax+ b)^2- 12x- 20(ax+ b)+ 36= 0. Solve that quadratic equation for x. Since the equation is quadratic, there may be two solutions giving the two points of intersection.
 
T
HallsofIvy said:
The center of the circle is (6, 10) and if it is tangent to the x-axis its radius is 10. The equation is (x- 6)^2+ (y- 10)^2= 100. That does reduce to x^2- 12x+ 36+ y^2- 20y+ 100= 100 or x^2+ y^2- 12x- 20y+ 36= 0 as you say.

Your are told that "the slope and the y-intercept of the straight line L are -1 and k respectively". Do you know what that means? Any (non-vertical) line can be written y= ax+ b. What are a and b?

To determine where the line intersects the circle, replace y in the equation of the circle with "ax+ b" (with the correct values for a and b of course) to get the equation x^2+ (ax+ b)^2- 12x- 20(ax+ b)+ 36= 0. Solve that quadratic equation for x. Since the equation is quadratic, there may be two solutions giving the two points of intersection.
Click to expand...
thanks for the idea.

But, I found out that the linear equation is y=-x+k, does it mean I have to sub this equation into the quadratic equation instead of finding x and y.

if yes, I cannot calculate further if it’s the case. Thanks again
 
nairb said:
T

thanks for the idea.

But, I found out that the linear equation is y=-x+k, does it mean I have to sub this equation into the quadratic equation instead of finding x and y.

if yes, I cannot calculate further if it’s the case. Thanks again
Click to expand...
Is there some other way to determine the intersection? Solve however you need to. Do what it takes to find the solution. It might be difficult and require care and focus.
 
tkhunny said:
Is there some other way to determine the intersection? Solve however you need to. Do what it takes to find the solution. It might be difficult and require care and focus.
Click to expand...
I tried to sub the linear equation into the circle equation, and let x = 0, and get y = 2 or y = 16. But after that, I’m not sure how can I find the mid point of A and B in terms of k. Thanks
 
nairb said:
I tried to sub the linear equation into the circle equation, and let x = 0, and get y = 2 or y = 16. But after that, I’m not sure how can I find the mid point of A and B in terms of k. Thanks
Click to expand...
Please show your work, if you really want help to correct it.
 
nairb said:
After substitution, I stuck here. Any ideas ?
Click to expand...
If you are so stuck then why do you post unreadable attachments?
Can you not learn to use LaTeX coding? After all, this is free help from professionals.
 
pka said:
If you are so stuck then why do you post unreadable attachments?
Can you not learn to use LaTeX coding? After all, this is free help from professionals.
Click to expand...
What is latex coding..？I know it is free help and I’m trying hard to figure out the question.
 
nairb said:
After substitution, I stuck here. Any ideas ?
Click to expand...


You failed to substitute y = - x + k everywhere there is a y.

Continuing:

\(\displaystyle x^2 + (-x + k)^2 - 12x - 20(-x + k) + 36 = 0\)

\(\displaystyle x^2 + x^2 - 2xk + k^2 - 12x + 20x - 20k + 36 = 0\)

\(\displaystyle 2x^2 + 8x - 2xk + k^2 - 20k + 36 = 0\)

Solve for x in terms of k, for example, by using the Quadratic Formula.

\(\displaystyle 2x^2 + (8 - 2k)x + (k^2 - 20k + 36) = 0\)

In this case for the Quadratic Formula, a = 2, \(\displaystyle \ \) b = (8 - 2k), and
c = (\(\displaystyle k^2 - 20k + 36\)).
 
Last edited:
lookagain said:


You failed to substitute y = - x + k everywhere there is a y.

Continuing:

\(\displaystyle x^2 + (-x + k)^2 - 12x - 20(-x + k) + 36 = 0\)

\(\displaystyle x^2 + x^2 - 2xk + k^2 - 12x + 20x - 20k + 36 = 0\)

\(\displaystyle 2x^2 + 8x - 2xk + k^2 - 20k + 36 = 0\)
Click to expand...
Yes I got the same result , but then how to find the coordinate of A and B In terms of k?
 
nairb said:
Yes I got the same result , but then how to find the coordinate of A and B In terms of k?
Click to expand...


Please look at the last part of the edited post #15 to get two x-values in terms of k
using the Quadratic Formula. I must go away now (to sleep).
 
lookagain said:


You failed to substitute y = - x + k everywhere there is a y.

Continuing:

\(\displaystyle x^2 + (-x + k)^2 - 12x - 20(-x + k) + 36 = 0\)

\(\displaystyle x^2 + x^2 - 2xk + k^2 - 12x + 20x - 20k + 36 = 0\)

\(\displaystyle 2x^2 + 8x - 2xk + k^2 - 20k + 36 = 0\)

Solve for x in terms of k, for example, by using the Quadratic Formula.

\(\displaystyle 2x^2 + (8 - 2k)x + (k^2 - 20k + 36) = 0\)

In this case for the Quadratic Formula, a = 2, \(\displaystyle \ \) b = (8 - 2k), and
c = (\(\displaystyle k^2 - 20k + 36\)).
Click to expand...
Wow, so the sum of root will be (-8+2k)/2, which is k-4. And sub (k-4) into the linear equation, and obtain y=2k-4.

So the mid point is [(k-4)/2, k-2]. Is that correct?

thanks so much !

 
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