Circle and Tangent Line

Definition of Circle Tangent Line

A circle tangent line is a line that intersects the circle at exactly one point. The intersection point between the tangent line and the circle is called the point of tangency.

Circle Tangent Line

A tangent line intersects the circle at exactly one point.

Important property: A tangent line is always perpendicular to the radius of the circle at the point of tangency.

Equation of Circle Tangent Line

Tangent Line Through a Point on the Circle

If point $(x_1, y_1)$ lies on the circle $x^2 + y^2 = r^2$ , then the equation of the tangent line at that point is:

x_1 \cdot x + y_1 \cdot y = r^2

For a circle with center $(a, b)$ :

(x_1 - a)(x - a) + (y_1 - b)(y - b) = r^2

Tangent Line with a Given Gradient

The equation of the tangent line to circle $x^2 + y^2 = r^2$ with gradient $m$ is:

y = mx \pm r\sqrt{1 + m^2}

For a circle with center $(a, b)$ :

y - b = m(x - a) \pm r\sqrt{1 + m^2}

Tangent Lines from an External Point

From a point outside the circle, two tangent lines can be drawn to the circle.

Two Tangent Lines from External Point

From point P outside the circle, two tangent lines can be drawn.

Length of Tangent Line

If $P(x_1, y_1)$ is a point outside the circle with center $O(a, b)$ and radius $r$ , then the length of the tangent line from P to the circle is:

PT = \sqrt{(x_1 - a)^2 + (y_1 - b)^2 - r^2}

Common Tangent Lines of Two Circles

External Common Tangent Lines

External common tangent lines are lines that touch both circles and do not intersect the line connecting the two circle centers.

External Common Tangent Lines

Lines that touch both circles from the same side.

Length of external common tangent line:

l = \sqrt{d^2 - (r_1 - r_2)^2}

where $d$ is the distance between the two circle centers.

Internal Common Tangent Lines

Internal common tangent lines are lines that touch both circles and intersect the line connecting the two circle centers.

Internal Common Tangent Lines

Lines that touch both circles from opposite sides.

Length of internal common tangent line:

l = \sqrt{d^2 - (r_1 + r_2)^2}

Determining Tangent Line Equations

Determining Tangent Line Through a Point on the Circle

Find the equation of the tangent line to circle $x^2 + y^2 = 25$ at point $(3, 4)$ .

Solution:

Since point $(3, 4)$ lies on the circle (can be verified: $3^2 + 4^2 = 9 + 16 = 25$ ), the equation of the tangent line is:

x_1 \cdot x + y_1 \cdot y = r^2

3x + 4y = 25

Determining Tangent Line with a Given Gradient

Find the equation of the tangent line to circle $x^2 + y^2 = 16$ that is parallel to line $3x - 4y + 5 = 0$ .

Solution:

The gradient of line $3x - 4y + 5 = 0$ is $m = \frac{3}{4}$ .

Equation of tangent line with gradient $m = \frac{3}{4}$ :

y = mx \pm r\sqrt{1 + m^2}

y = \frac{3}{4}x \pm 4\sqrt{1 + \left(\frac{3}{4}\right)^2}

y = \frac{3}{4}x \pm 4\sqrt{1 + \frac{9}{16}}

y = \frac{3}{4}x \pm 4\sqrt{\frac{25}{16}}

y = \frac{3}{4}x \pm 4 \cdot \frac{5}{4}

y = \frac{3}{4}x \pm 5

Therefore, the equations of the tangent lines are:

$y = \frac{3}{4}x + 5$ or $3x - 4y + 20 = 0$
$y = \frac{3}{4}x - 5$ or $3x - 4y - 20 = 0$

Calculating the Length of Tangent Line from External Point

Find the length of the tangent line from point $P(7, 1)$ to circle $x^2 + y^2 = 25$ .

Solution:

Circle center $O(0, 0)$ and radius $r = 5$ .

PT = \sqrt{(x_1 - a)^2 + (y_1 - b)^2 - r^2}

PT = \sqrt{(7 - 0)^2 + (1 - 0)^2 - 25}

PT = \sqrt{49 + 1 - 25}

PT = \sqrt{25}

PT = 5

Practice Problems

Find the equation of the tangent line to circle $x^2 + y^2 = 36$ at point $(-3, 3\sqrt{3})$ !
Find the equation of the tangent line to circle $(x - 2)^2 + (y + 3)^2 = 25$ that is perpendicular to line $2x + y - 5 = 0$ !
From point $A(10, 0)$ tangent lines are drawn to circle $x^2 + y^2 = 36$ . Find:
- Length of tangent line
- Coordinates of tangent points
Two circles are centered at $O_1(-4, 0)$ with radius 2 and $O_2(4, 0)$ with radius 3. Find the length of the external common tangent line!
Find the equation of the tangent line to circle $x^2 + y^2 - 6x + 4y - 12 = 0$ that passes through point $(8, -2)$ !

Answer Key

Tangent line equation at a point on the circle

Verify point on circle:

$(-3)^2 + (3\sqrt{3})^2 = 9 + 27 = 36 \space \checkmark$

Tangent line equation:

$x_1 \cdot x + y_1 \cdot y = r^2$
$-3x + 3\sqrt{3}y = 36$
$-x + \sqrt{3}y = 12$
$x - \sqrt{3}y + 12 = 0$
Tangent line perpendicular to a given line

Gradient of line $2x + y - 5 = 0$ is $m_1 = -2$ .

Since perpendicular, then $m_2 = \frac{1}{2}$ .

Tangent line equation:

$y - b = m(x - a) \pm r\sqrt{1 + m^2}$
$y + 3 = \frac{1}{2}(x - 2) \pm 5\sqrt{1 + \frac{1}{4}}$
$y + 3 = \frac{1}{2}(x - 2) \pm 5 \cdot \frac{\sqrt{5}}{2}$
$y + 3 = \frac{1}{2}x - 1 \pm \frac{5\sqrt{5}}{2}$

Therefore: $x - 2y - 4 \pm 5\sqrt{5} = 0$
Tangent line from external point
- Length of tangent line:
  
  $AT = \sqrt{10^2 + 0^2 - 36} = \sqrt{100 - 36} = \sqrt{64} = 8$
- Coordinates of tangent points can be found using the tangent line equation from external point.
External common tangent line

$d = \sqrt{(4-(-4))^2 + (0-0)^2} = 8$
$l = \sqrt{d^2 - (r_2 - r_1)^2} = \sqrt{64 - 1} = \sqrt{63} = 3\sqrt{7}$
Tangent line through external point

Circle: $(x - 3)^2 + (y + 2)^2 = 25$

Center $(3, -2)$ , radius $r = 5$

Verify point $(8, -2)$ outside circle:

$(8-3)^2 + (-2+2)^2 = 25 + 0 = 25$

Point is exactly on the circle! Therefore the tangent line equation is:

$(8-3)(x-3) + (-2+2)(y+2) = 25$
$5(x-3) + 0 = 25$
$x - 3 = 5$
$x = 8$

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