Maths > Circles > 8.0 Equation of tangent to a circle

  Circles
    1.0 Definition
    2.0 Equation of circle in various forms
    3.0 Intercepts made by a circle on coordinate axis
    4.0 Position of a point with respect to a circle
    5.0 Maximum and minimum distance of a point from a circle
    6.0 Intersection of a line and a circle
    7.0 Length of intercept cutoff from a line by a circle or length of chord of a circle
    8.0 Equation of tangent to a circle
    9.0 Tangents from a point to the circle
    10.0 Length of tangent from a point to a circle
    11.0 Common Tangents
    12.0 Equation of common tangents
    13.0 Pair of tangents
    14.0 Normal to a circle at a given point
    15.0 Common chord of two circles
    16.0 Equation of chord joining two points on circle
    17.0 Equation of chord of circle whose midpoint is given
    18.0 Chord of contact
    19.0 Orthogonal Circles
    20.0 Director Circle
    21.0 Family of circles

8.1 Slope form
This is used when the slope of tangent is given. Let us consider a equation of tangent $y = mx + c'$ to a circle ${x^2} + {y^2} + 2gx + 2fy + c = 0$.

Either value of slope $m$ of tangent is given directly or using the data given in the question we have to find out the slope $m$ of the tangent.

After finding the slope, using condition of tangency, we can find the value of constant $c'$ in the equation of tangent i.e.,

Radius of circle$=$Perpendicular distance of centre of a circle to the line/tangent of circle.

The equation of a tangent of slope $m$ to the circle ${x^2} + {y^2} = {a^2}$ is $y = mx \pm a\sqrt {1 + {m^2}} $ and the coordinates of point of contact are $\left( { \pm \frac{{am}}{{\sqrt {1 + {m^2}} }}, \mp \frac{a}{{\sqrt {1 + {m^2}} }}} \right)$.

Proof: Let $y=mx+c$ is the tangent of circle ${x^2} + {y^2} = {a^2}$.
Therefore, length of perpendicular from centre of circle $(0,0)$ on $y=mx+c$$=$ radius of circle.
$$\left| {\frac{c}{{\sqrt {1 + {m^2}} }}} \right| = a$$ or, $$c = \pm a\sqrt {1 + {m^2}} $$
Substituting this value of $c$ in $y=mx+c$, we get $y = mx \pm a\sqrt {1 + {m^2}} $ which are the required equations of tangents.


Question 15. Find the equation of tangent to the circle ${x^2} + {y^2} - 2x - 2y - 1 = 0$ such that it makes an angle of ${60^ \circ }$ with $X$-axis.

Solution: From the equation of circle, centre of circle is $(1,1)$ and radius $r = \sqrt {{g^2} + {f^2} - c} = \sqrt { - {1^2} + - {1^2} + 1} = \sqrt 3 $.
Since the tangent makes an angle of ${60^ \circ }$ with $X$-axis $$m = \tan 60^\circ = \sqrt 3 $$
The equation of tangent is $$y = \sqrt 3 x + c$$
Now, the length of perpendicular drawn from centre of circle to the tangent $=$radius of circle
$$r = \left| {\frac{{1 - \sqrt 3 - c}}{{\sqrt {3 + 1} }}} \right|$$ $$\sqrt 3 = \left| {\frac{{1 - \sqrt 3 - c}}{2}} \right|$$ $$ \pm 2\sqrt 3 = 1 - \sqrt 3 - c$$ $$c = 1 - 3\sqrt 3 $$ or, $$c = 1 + \sqrt 3 $$
The equation of tangents are $y = \sqrt 3 x + 1 - 3\sqrt 3 $ and $y = \sqrt 3 x + 1 + \sqrt 3 $.
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