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Tangent lines from a point to a circle calculator

Print tangent lines from a point to circle calculator
Circles form:   (x − a)2 + (y − b)2 = r2 ( x - )2 + ( y - )2 = 2
Circles form: x2 + y2 + Ax + By + C = 0 x2 + y2 + x + y + = 0
Point outside the circle:     input (x, y): ( , )
Circle equation:
Tangent line 1 Equation:
Tangency point:
Tangent line 2 Equation:
Tangency point:
Distance from point to circle center:
Line connecting circle center to the point Line equation:
Line angle: Degree
x axis intercept:
y axis intercept:
     
Tangent lines summary
Circle equations summary
Line basic definitions
Example 1
Example 2

Tangent lines between circle and a point summary

Print circle point tangent lines summary
Scheme of tangent line to circle
Tangent lines to a circle
The distance between the point (xp , yp) and the tangent point (1) is:

The angle between the two tangent lines   θ   is:

Note: in the equations above x1 can be replaced by x2 (they are equal).

Circle form, center at (0 , 0): x2 + y2 = r2

Tangents x and y points:

(1)
(2)
Distance from (xp , yp) To circle center:
Equation of the line connecting point (xp , yp) with circle center:
Circle form, center at (h , k): (x − h)2 + (y − k)2 = r2
Tangent x points:
(3)
Tangents y points:
(4)
Distances:
Equation of the line connecting point (xp , yp) with circle center:
y=(b-y_p)/(a-x_p ) x-(x_p (b-y_p ))/(a-x_p )+y_p
x and y intercepts:
x_i=(ay_p-bx_p)/(x_p-a)
y_i=(x_p (b-y_p ))/(x_p-a)+y_p
Circle form: x2 + y2 + Ax + By + C = 0
x_(1,2)=(2r^2 (2x_p+A)±r(2y_p+B) √((2x_p+A)^2+(2y_p+B)^2-4r^2 ))/((2x_p+A)^2+(〖2y〗_p+B)^2 )-A/2
(5)
y_(1,2)=(2r^2 (2y_p+B)∓r(2x_p+A) √((2x_p+A)^2+(2y_p+B)^2-4r^2 ))/((2x_p+A)^2+(〖2y〗_p+B)^2 )-B/2
(6)
D=1/2 √((2x_p+A)^2+(2y_p+B)^2 )

Equation of the line connecting point (xp , yp) with circle center

Equation:
y=(2y_p+B)/(2x_p+A) x-a(2y_p+B)/(2x_p+A)-B/2
x intercept:
x_i=(Bx_p-Ay_p)/(2y_p+B)
y intercept:
y_i=y_p-(x_p (B+2y_p ))/(2x_p+A)
If two points (x1 , y1) and (x2 , y2) on the circle are given then the intersection point (xp , yp ) created by the tangents lines is:
Where:
and

Example 1 - Circle tangent lines

Print example 1
Find the equations of the line tangent to the circle given by:  x2 + y2 + 2x − 4y = 0  at the point  P(1 , 3).
Tangent line
The incline of a line tangent to the circle can be found by implicit derivation of the equation of the circle related to x (derivation dx / dy)
If the equation of the circle is:
Ax2 + Ay2 + Bx + Cy + D = 0
Then the implicit derivation is:
2Ax+2Ay dy/dx+B+C dy/dx=0
dy/dx=(-2Ax-B)/(2Ay+C)
dy/dx=m=(-2∙1∙1-2)/(2∙1∙3-4)=-2
The slope of the tangent line can be expressed by the points (x , y) and P as the tangent of  θ  or  m  as:
m=(y-3)/(x-1) (tangent of the line)
y − 3 = mx − m mx − y + 3 − = 0
And the equation of the tangent line is:      2x + y − 5 = 0

Example 2 - Tangent lines to a circle

Print example 2
Find the equations of the tangent lines to the circle given by the equation:  x2 + y2 + 4x + 6y − 21 = 0  from the point  P(−4 , 5).
Example 2 - general tangency of two lines and a circle
A simple way to find the points S and Q is by applying geometric methods.
The distance from the point to the circle center is:
(PO) ̅=√((x_p-x_c )^2+(y_p-y_c )^2 )
Example 2 - two tangents lines
First, we shell find the center and the radius of the circle, by the procedure for completing the square:
(x^2+4x)+(y^2+6y)-21=0
(x+2)^2-4+(y+3)^2-9-21=0
(x+2)^2+(y+3)^2=34
We can see that the center of the circle is at  (−2 , −3)  and the radius is equal to:  √34.
Since point  Q  is common to the line and the circle, its  x  and  y  coordinates are the same and we have to solve the set of equations:
Ax+By+C=0 (x-c)^2+(y-k)^2=r^2
From the line equation after eliminating y we get:
y=(-Ax-C)/B
substitute y into the circle equation.
(x-c)^2+(-A/B x-(C+Bk)/B)^2=r^2
x^2-2cx+c^2-A^2/B^2 x^2-2A(C+Bk)/B x-(4A^2)/B^2 (C+Bk)^2-r^2=0
x^2 (1-A^2/B^2 )-x(2c-2A(C+Bk)/B)+c^2-(4A^2)/B^2 (C+Bk)^2-r^2=0

this is a quadratic equation for  x: