Circles
1.0 Definition
2.0 Equation of circle in various forms
2.1 Central Form
2.2 Standard Form
2.3 Parametric Form
2.4 General form of circle
2.4.1 Application of general form of circle
2.5 Diametric form of circle
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
8.1 Slope form
8.2 Point form
8.3 If a point outside the circle is given through which tangent to a circle passes, then
8.4 Parametric form
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
8.3 If a point outside the circle is given through which tangent to a circle passes, then
2.2 Standard Form
2.3 Parametric Form
2.4 General form of circle
2.4.1 Application of general form of circle
2.5 Diametric form of circle
8.2 Point form
8.3 If a point outside the circle is given through which tangent to a circle passes, then
8.4 Parametric form
$(a)$ Assume equation of line passing through point $P({x_1},{y_1})$ with slope $m$ as $$\begin{equation} \begin{aligned} y - {y_1} = m\left( {x - {x_1}} \right) \\ y - mx + m{x_1} - {y_1} = 0 \\\end{aligned} \end{equation} $$
$(b)$ Perpendicular distance from centre of circle to the above line is equal to radius of circle since it is a tangent and find the value of $m$ using the quadratic equation formed which gives two values of $m$.
Question 18. Find the equation of tangent to the circle ${x^2} + {y^2} = 16$ which passes through the point $(4,5)$.
Solution: Let us assume the slope of tangent passing through point $P(4,5)$ be $m$. Equation of tangent in terms of slope is $$y - 5 = m\left( {x - 4} \right)$$ $$y - mx + 4m - 5 = 0$$
As the centre of circle is origin $O(0,0)$ and radius $r=4$.
Length of perpendicular from centre of circle to the tangent$ = \left| {\frac{{4m - 5}}{{\sqrt {1 + {m^2}} }}} \right|$, which is equal to the radius of circle i.e., $\left| {\frac{{4m - 5}}{{\sqrt {1 + {m^2}} }}} \right| = 4$
Squaring both sides, we get $${(4m - 5)^2} = 16(1 + {m^2})$$ $$16{m^2} + 25 - 40m = 16 + 16{m^2}$$
Therefore, $$40m=9$$ or, $$m = \frac{9}{{40}}$$ and $$m = \infty {\text{ }}(as{\text{ }}highest{\text{ }}power{\text{ }}of\ m{\text{ }}cancel{\text{ }}each{\text{ }}other)$$
Therefore, the equation of tangents are $$x=4$$ and $$y - 5 = \frac{9}{{40}}\left( {x - 4} \right)$$
