Tangent Line To A Circle

mathxyz · 2024-03-02 04:29:08

The tangent line to a circle may be defined as the line that intersects the circle in a single point called the point of tangency. If the equation of the circle is x^2 + y^2 = r^2 and the equation of the tangent line is y = mx + b, show that:

A. r^2(1 + m^2) = b^2

HINT GIVEN IN THE TEXTBOOK:

The quadratic equation x^2 + (mx + b)^2 = r^2 has exactly one solution.

B. The point of tangency is [(-r^2•m)/b, (r^2)/b].

C. The tangent line is perpendicular to the line containing the center of the circle and the point of tangency.

I need help with all three parts of this question.

Bob · 2024-03-02 05:09:04

If you substitute ( x , mx+c) into the circle equation you get that hint.

A quadratic Ax^2 + Bx + C = 0 has only one solution if B^2 - 4AC is zero. Part 1 comes from that equation substituting for A, B and C. (Don't mix up C and c here; they're not the same.)

If P is the point of tangency and origin O then OP is perpendicular to the tangent so it has equation y = -x/m

Sub into the equation for the circle to find x and then y.

What is the question for part c?

Bob

mathxyz · 2024-03-02 18:47:20

Bob wrote:

If you substitute ( x , mx+c) into the circle equation you get that hint.
A quadratic Ax^2 + Bx + C = 0 has only one solution if B^2 - 4AC is zero. Part 1 comes from that equation substituting for A, B and C. (Don't mix up C and c here; they're not the same.)
If P is the point of tangency and origin O then OP is perpendicular to the tangent so it has equation y = -x/m
Sub into the equation for the circle to find x and then y.
What is the question for part c?
Bob

Part C

Show that the tangent line is perpendicular to the line containing the center of the circle and the point of tangency.

mathxyz · 2024-03-02 18:58:29

Bob wrote:

If you substitute ( x , mx+c) into the circle equation you get that hint.
A quadratic Ax^2 + Bx + C = 0 has only one solution if B^2 - 4AC is zero. Part 1 comes from that equation substituting for A, B and C. (Don't mix up C and c here; they're not the same.)
If P is the point of tangency and origin O then OP is perpendicular to the tangent so it has equation y = -x/m
Sub into the equation for the circle to find x and then y.
What is the question for part c?
Bob

You said:

"If you substitute ( x , mx+c) into the circle equation you get that hint."

Which equation of the circle?

You said:

"A quadratic Ax^2 + Bx + C = 0 has only one solution if B^2 - 4AC is zero. Part 1 comes from that equation substituting for A, B and C. (Don't mix up C and c here; they're not the same.)"

What do you mean here?

You said:

If P is the point of tangency and origin O then OP is perpendicular to the tangent so it has equation y = -x/m."

I don't know what this equation represents.

You said:

"Sub into the equation for the circle to find x and then y."

Which equation of the circle?

Last edited by mathxyz (2024-03-02 18:59:13)

KerimF · 2024-03-03 04:36:54

FelizNYC wrote:

the equation of the circle is x^2 + y^2 = r^2 and the equation of the tangent line is y = mx + b

Please note that the following simple steps don't need knowing the hint.

In general, when we have two traces (here, a circle and a straight line) and we like to find at which point(s) they may intersect, we simply solve their two equations.

We have:
x^2 + y^2 = r^2
y = mx + b
And their unknown variables are x and y.

By substituding y, we get:
x^2 + (mx + b)^2 = r^2
x^2 + m^2*x^2 + 2*m*x*b + b^2 = r^2
(1 + m^2)*x^2 + 2*m*b*x + (b^2 - r^2) = 0

This is a quadratic equation in the form of:
A*x^2 + B*x + C =0
where
A = (1 + m^2)
B = 2*m*b
C = (b^2 - r^2)

Its solution is:
x1 = [-B - sqrt(delta)]/(2*A)
x2 = [-B + sqrt(delta)]/(2*A)

Where delta is:
B^2 - 4*A*C

Therefore the delta of our equation above is:
= B^2 - 4*A*C
= (2*m*b)^2 - 4*(1 + m^2)*(b^2 - r^2)
= 4*m^2*b^2 - 4*(b^2 - r^2 + m^2*b^2 - m^2*r^2)
= 4*(m^2*b^2 - b^2 + r^2 - m^2*b^2 + m^2*r^2)
= 4*(- b^2 + r^2 + m^2*r^2)
= 4*[r^2*(1 + m^2) - b^2]

Since the straight line intersects the circle at one point only, x1 and x2 of this equation should be equal.
They are equal if its delta=0, that is:
4*[r^2*(1 + m^2) - b^2] = 0
r^2*(1 + m^2) - b^2 = 0
r^2*(1 + m^2) = b^2

Last edited by KerimF (2024-03-03 04:38:29)

mathxyz · 2024-03-03 18:08:44

KerimF wrote:

FelizNYC wrote:
the equation of the circle is x^2 + y^2 = r^2 and the equation of the tangent line is y = mx + b
Please note that the following simple steps don't need knowing the hint.
In general, when we have two traces (here, a circle and a straight line) and we like to find at which point(s) they may intersect, we simply solve their two equations.
We have:
x^2 + y^2 = r^2
y = mx + b
And their unknown variables are x and y.
By substituding y, we get:
x^2 + (mx + b)^2 = r^2
x^2 + m^2*x^2 + 2*m*x*b + b^2 = r^2
(1 + m^2)*x^2 + 2*m*b*x + (b^2 - r^2) = 0
This is a quadratic equation in the form of:
A*x^2 + B*x + C =0
where
A = (1 + m^2)
B = 2*m*b
C = (b^2 - r^2)
Its solution is:
x1 = [-B - sqrt(delta)]/(2*A)
x2 = [-B + sqrt(delta)]/(2*A)
Where delta is:
B^2 - 4*A*C
Therefore the delta of our equation above is:
= B^2 - 4*A*C
= (2*m*b)^2 - 4*(1 + m^2)*(b^2 - r^2)
= 4*m^2*b^2 - 4*(b^2 - r^2 + m^2*b^2 - m^2*r^2)
= 4*(m^2*b^2 - b^2 + r^2 - m^2*b^2 + m^2*r^2)
= 4*(- b^2 + r^2 + m^2*r^2)
= 4*[r^2*(1 + m^2) - b^2]
Since the straight line intersects the circle at one point only, x1 and x2 of this equation should be equal.
They are equal if its delta=0, that is:
4*[r^2*(1 + m^2) - b^2] = 0
r^2*(1 + m^2) - b^2 = 0
r^2*(1 + m^2) = b^2

Nicely-done. Very detailed. Very informative. I thank you.

Math Is Fun Forum

#1 2024-03-02 04:29:08

Tangent Line To A Circle

#2 2024-03-02 05:09:04

Re: Tangent Line To A Circle

#3 2024-03-02 18:47:20

Re: Tangent Line To A Circle

#4 2024-03-02 18:58:29

Re: Tangent Line To A Circle

#5 2024-03-03 04:36:54

Re: Tangent Line To A Circle

#6 2024-03-03 18:08:44

Re: Tangent Line To A Circle

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