In the case of the circle, the normal is a radius.
First, we cover how to find the equation of a tangent to a circle when we know the point of contact on the circle's circumference.
Second, we cover how to find the equations of the tangents (notice the plurals) to a circle when we know the coordinates of an external point and the equation of the circle.
In this case we find two linear equations because there are always 2 tangents to any circle from any external point.
An external point is one that is not on or within the circle's circumference.
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| intro | given point of contact | from external point |
| examples | practice | solutions |
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Equations of Tangents to Circles Given the Point of Contact
Because we know that the radius of a circle is perpendicular to the tangent at the point of contact, and we also know that perpendicular lines have negative reciprocal slopes, it is fairly easy to find the equation of a tangent to a circle when given the point of contact.
Since we know the coordinates of the center of the circle, and we're given the coordinates of the point of contact of the tangent, we need only make a statement that says that the slope between any point P(x, y), on the tangent and the point of contact on the circumference equals the negative reciprocal of the slope between the center of the circle and the point of contact of the tangent.
Simply put, write an equation that says:
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| Tangent slope = negative reciprocal of radius slope. |
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The variable in our equation is the slope of the tangents. We will get 2 answers.
Let's do an example and then we'll generalize the process.
We'll start with a circle center at the origin, then apply what we learn to any circle.
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Example
Find the equation of the tangent to the circle x2 + y2 = 10
which contacts the circle at the point (1, 3).
Solution:
The slope of the tangent is:
The slope of the radius is 3.
Since the tangent is z the radius, the tangent
has a slope of -1/3
‹ 
Had the center of our circle been at a point other than the origin, the only change in the process would be that the slope of the radius would not have been quite so simple to find, however, the approach would be exactly the same.
Our statement would still say the slope between any point on the tangent and the point of contact is equal to the negative reciprocal of the slope of the radius to the point of contact.
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| The equation of the tangent which touches circle center (h, k) at the point (x1, y1) is y - y1 = |
Though we have a formula for finding the equation of the tangent, if we do each question individually and simply write an equation that says that the slope of the tangent is the negative reciprocal of the slope of the radius to the point of contact we don't have to worry about forgetting the formula. Remember the logic rather than the formula -- it's easier!
Let's do another example, this one with the center at a point other than the origin.
Example
Find the equation of the tangent to the circle x2 + y2 + 4x - 8y - 5 = 0 at P(2, 1).
Solution:
The center of the circle is (-2, 4). The slope of the radius from (2, 1) is -3/4.
The slope of the tangent is 4/3 so the equation is 
4x - 3y - 5 = 0.
Important Note:
The equation of a circle must have both x2 and y2 terms in it and they both must have the same coefficient. In all the examples here, that coefficient is 1 -- however, it need not be. The coefficients of x2 and y2 can be any real number, but they must be equal.
This means that 3x2 + 3y2 - 12x + 18y - 9 = 0 is the equation of a circle. In order to solve for the center and radius however, we must put the equation in general form -- that is -- we must divide through by 3 in order to make the coefficients of x2 and y2 equal to 1.
So, the equation mentioned above becomes:
x2 + y2 - 4x + 6y - 3 = 0
and we can use the techniques discussed earlier to find the center and radius.
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| intro | given point of contact | from external point |
| examples | practice | solutions |
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Equation of a Tangent to a Circle from an External Point
Example
Find the equation(s) of the tangent(s) from P(-1, 7) to the circle x 2 + y 2 = 5.
If m represents the slope of PA, since P is (-1, 7), we can say that
or y - 7 = m(x + 1).
This means that the equation of PA is: mx - y + (m + 7) = 0.
We know that the distance from the origin to A = the radius =
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so the distance from (0, 0) to mx - y + (m + 7) = 0 is 
, the radius.
Using our formula for perpendicular distance from a point to a line,
the distance from (0, 0) to the line mx - y + (m + 7) = 0 is:
When we simplify this equation we get: 2m2 - 7m - 22 = 0 from which we get that:
m = -2 or m = 
When we substitute these two values for m into mx - y + (m + 7) = 0, the equation for PA
we get: -2x - y + 5 = 0 for m = -2 and 11x - 2y + 25 = 0 for m = 
The first equation, in General Form 2x + y - 5 = 0, is PA and the second is PB. (diagram)
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To find the equation of a tangent to a circle center O from an external point P we:
| 1) Write an equation for the line through P with slope = m. 2) Write an equation stating that the perpendicular distance from O to the tangent equals the radius of the circle. 3) Solve this equation for m, substitute the solutions into the equation in step 1. |
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| intro | given point of contact | from external point |
| examples | practice | solutions |
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Example
Find the equation of the tangent from A(2, -3) to the circle x2 + y2 + 2x + 4y + 3 = 0.
Solution:
First, we must find the center and radius of the circle:
the center C is C(-1, -2)
the radius is found from c = h2 + k2 - r2 or 3 = (-1)2 + (-2)2 - r2 u r = 
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Since the tangent goes through A, its equation is 
or mx - y + (-3 - 2m) = 0.
Now we write an equation that states that the perpendicular distance from C(-1, -2)
to the line mx - y + (2 - m) = 0 equals the radius:
7m2 + 6m - 1 = 0 u (7m - 1)(m + 1) = 0
m = -1 or m = 1/7.
When m = -1, mx - y + (-3 - 2m) = 0 becomes x + y + 1 = 0
When m = 1/7, mx - y + (-3 - 2m) = 0 becomes x - 7y - 23 = 0.
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Example
Find the points of contact of the two tangents in the previous example .
Solution:
Now, we have to find the coordinates of the points common to the tangents and the circle.
So, we have to solve a system of mixed equations.
Recall that when we solve a linear and a quadratic equation simultaneously, we use the linear equation to generate a substitution expression for one of the variables.
To find the first point of tangency the system to solve is:
x2 + y2 + 2x + 4y + 3 = 0
x + y + 1 = 0 .
x + y + 1 = 0 u y = -x - 1
substituting for y, we get:
x2 + (-x - 1)2 + 2x + 4(-x - 1) + 3 = 0 u 2x2 = 0 u x = 0 u y = -1
The point of contact is (0, -1).
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For the other tangent, the system to solve is:
x2 + y2 + 2x + 4y + 3 = 0
x - 7y - 23 = 0
x - 7y - 23 = 0 u x = 7y + 23
and substituting for x, we get:
(7y + 23)2 + y2 + 2(7y + 23) + 4y + 3 = 0 u
25y2 + 170y + 289 = 0
so y = -17/5.
Substituting into x - 7y - 23 = 0 we get
x = -4/5 making the point of contact 
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Example
Find the equations of the tangents from the origin to the circle center (1, -2), radius = 1.
Solution: (make a diagram on graph paper please!)
The equation of a line through (0, 0) with slope m is: y = mx u mx - y = 0.
The perpendicular distance from (1, -2) to mx - y = 0 is 
We set the distance = 1 and solve for m:
m 2 + 4m + 4 = m 2 + 1 u m = -3/4
so 3x + 4y = 0 is the equation of one tangent.
Now, what about the other tangent?
If we look at a picture of this setup, we see the center at (1, -2) and the radius = 1,
so the y-axis is tangent to the circle since it is exactly one unit away from the center.
(look at your diagram!)
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The equation of this tangent then is x = 0.
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There are 2 tangents to any circle from any external point.
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(look at your diagram!!!)
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Example
Given point P(-3, -2) and C, a circle with equation x2 + y2 - 4x - 6y + 8 = 0, find:
(a) The equation of the tangents from P to C,
(b) The coordinates of the points of contact,
(c) The equations of the radii to the points of contact.
Solution: (make a diagram!!)
(a) C's center is (2, 3), the radius = 
The equation of a line through P with slope m is 
mx - y + (3m - 2) = 0
The perpendicular distance from (2, 3) to mx - y + (3m - 2) = 0 is:
and it must = 
So, solving 
we get
2m2 - 5m + 2 = 0 u (2m - 1)(m - 2) = 0 u m = ½ or m = 2
The tangent equations are: x - 2y - 1 = 0 for m = ½
2x - y + 4 = 0 for m = 2.
(b) For the tangent x - 2y - 1 = 0, the coordinates of the point of contact are:
We substitute x = 2y + 1 into x2 + y2 - 4x - 6y + 8 = 0 and solve.
(2y + 1)2 + y2 - 4(2y + 1) - 6y + 8 = 0 u y2 - 2y + 1 = 0 u y = 1, x = 3
The point of contact is (3, 1).
For the tangent x = 0, we simply set x = 0 in 2x - y + 4 = 0 to get y = 4.
The point of contact is (0, 4).
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(c) The equation of the radius from the center (2, 3) to (3, 1) is:
2x + y - 7 = 0.
The equation of the radius from the center (2, 3) to (0, 4) is:
x + 2y - 8 = 0
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| intro | given point of contact | from external point |
| examples | practice | solutions |
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Make Diagrams for numbers 4, 5 and 6!!
1) Write the equation of the tangent to the circle x 2 + y 2 = 65 at (-4, 7).
2) Write the equation of the tangent to the circle x 2 + y 2 - 8y - 37 = 0 at (-7, 6).
3) Write the equation of the tangent to the circle x2 + y2 - 2x + y - 1 = 0 at (1, -2)
4) Find the equations of the tangents to the given circle from the given external point:
(a) x 2 + y 2 = 25; (7, 5)
(b) x 2 + y 2 = 20; (-4, -3)
(c) x 2 + (y + 2) 2 = 2; (1, 1)
5) Find the equations of the tangents to the circle x 2 + y 2 = 5 from the point (3, -1).
6) Find the equations of the tangents to x 2 + y 2 - 4x - 6y + 11 = 0 from the point (5, 8).
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| intro | given point of contact | from external point |
| examples | practice | solutions |
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1) Write the equation of the tangent to the circle x 2 + y 2 = 65 at (-4, 7).
The center is (0,0), so the slope of the radius to (-4, 7) = -7/4
So the slope of the tangent = 4/7 u the equation of the tangent is 
When we write this in Ax + By + C = 0 form, we get: 4x - 7y + 65 = 0.
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2) Write the equation of the tangent to the circle x 2 + y 2 - 8y - 37 = 0 at (-7, 6).
The center is (0, 4), so the slope of the radius to (-7, 6) = -2/7
‹ the slope of the tangent is 7/2 u the equation of the tangent is 
When we write this in Ax + By + C = 0 form, we get: 7x - 2y + 61 = 0.
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3) Write the equation of the tangent to the circle x 2 + y 2 - 2x + y - 1 = 0 at (1, -2).
The center is (1, -½) so the slope of the radius to (1, -2) = undefined.
This means that the tangent is y the x-axis, since the radius is y the y-axis.
The equation of the tangent therefore is y = -2.
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4) Find the equations of the tangents to the given circle from the given external point:
| CIRCLE | POINT | TANGENT -1 | TANGENT -2 |
| a) x 2 + y 2 = 25; | (7, 5) | y = 5 | 35x - 12y - 185 = 0 |
| b) x 2 + y 2 = 20 | (-4, -3) | x + 2y + 10 = 0 | 11x + 2y + 50 = 0 |
| c) x 2 + (y + 2) 2 = 2 | (1, 1) | x - y = 0 | x + 7y - 8 = 0 |
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5) Find the equations of the tangents to the circle x2 + y2 = 5 from the point (3, -1).
Tangent-1: 2x + y - 5 = 0,
Tangent-2: x - 2y - 5 = 0
6) Find equations of the tangents to the circle x2 + y2 - 4x - 6y + 11 = 0 from (5, 8).
Tangent-1: x - y + 3 = 0
Tangent-2: 23x - 7y - 59 = 0
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| intro | given point of contact | from external point |
| examples | practice | solutions |
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