If the point of contact between a circle and a tangent is known, then the equation of the tangent can be calculated.
If a line is a tangent to a circle, then a radius will meet the tangent at right angles. The gradient of this radius can be calculated, since the center and point of contact are known.
Using radius tangent
m _{radius} × m _{tangent} = −1
the gradient of the tangent can be found.
The equation can then be found using
y − y_{1} = m(x − x_{1})
since the point is known, and the gradient has just been calculated.
Problem 1 :
Show that A (1, 3) lies on the circle x^{2} + y^{2} + 6x + 2y − 22 = 0 and find the equation of the tangent at A.
Solution :
To show that the given point lies on the circle, let us apply the point into the equation of circle.
If x = 1 and y = 3
1^{2} + 3^{2} + 6(1) + 2(3) − 22 = 0
1 + 9 + 6 + 6 - 22 = 0
22 - 22 = 0
0 = 0
Since the given point satisfies the equation of the circle, the point must lie on the circle.
Finding center of the circle :
x^{2} + y^{2} + 6x + 2y − 22 = 0
x^{2} + 6x + y^{2} + 2y − 22 = 0
(x + 3)^{2} + (y + 1)^{2} - 3^{2} - 1^{2} - 22 = 0
(x + 3)^{2} + (y + 1)^{2} = 32
Center of the circle is (-3, -1).
Slope of radius from (-3, -1) and (1, 3) :
m = (y_{2} - y_{1})/(x_{2} - x_{1})
m = (3+1)/(1+3)
m = 4/4
m = 1
Slope of tangent = -1/Slope of radius
= -1
Equation of tangent :
(y - y_{1}) = m(x - x_{1})
Equation of the tangent at the point (1, 3).
y - 3 = -1(x - 1)
y - 3 = -x + 1
x + y = 1 + 3
x + y = 4
So, equation of tangent for the given circle is x + y = 4.
Problem 2 :
Find the equation of the tangent to the circle
x^{2} + y^{2} - 3x + y - 10 = 0
through the point R (-1, 2) on its circumference.
Solution :
x^{2} + y^{2} - 3x + y - 10 = 0
Finding center of the circle :
x^{2} - 3x + y^{2}+ y - 10 = 0
(x - 3/2)^{2} - (3/2)^{2} + (y + 1/2)^{2} - (1/2)^{2} - 10 = 0
(x - 3/2)^{2} + (y + 1/2)^{2} - (9/4) - (1/4) - 10 = 0
(x - 3/2)^{2} + (y + 1/2)^{2} = (50/4)
Center (3/2, -1/2)
Slope of the radius from (3/2, -1/2) and R(-1, 2) :
m = (2 + 1/2)/(-1 - 3/2)
m = (5/2) / (-5/2)
m = -1
Slope of the tangent = -1/(-1) ==> 1
Equation of the tangent line :
Point (-1, 2) and slope = 1
y - 2 = 1(x + 1)
y - 2 = x + 1
x - y = -2 - 1
x - y = -3
Problem 3 :
Find the equation of tangent to the circle
x^{2} + y^{2} + 4x + 6y - 5 = 0
from the point R(1, 6) on its circumference.
Solution :
x^{2} + y^{2} + 4x + 6y - 5 = 0
Finding center of the circle :
x^{2} + 4x + y^{2 }+ 6y - 5 = 0
(x + 2)^{2} + (y + 3)^{2} - 2^{2} - 3^{2} - 5 = 0
(x + 2)^{2} + (y + 3)^{2} = 18
Center (-2, -3)
Slope of the radius from (-2, -3) and R(1, 6) :
m = (6 + 3)/(1 + 2)
m = 9 / 3
m = 3
Slope of the tangent = -1/3
Equation of the tangent line :
Point (1, 6) and slope = 1/3
y - 6 = (1/3)(x - 1)
3(y - 6) = x - 1
3y - 18 = x - 1
x - 3y = -18 + 1
x - 3y = -17
So, equation of the tangent is x - 3y = -17.
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM