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
8.2 Point 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
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
This is used to find the tangent at a point which lies on a circle. Equation of circle is ${x^2} + {y^2} + 2gx + 2fy + c = 0$. To find the tangent at point $P({x_1},{y_1})$, we use a standard form $T=0$ which can be written as $$x{x_1} + y{y_1} + g\left( {x + {x_1}} \right) + f\left( {y + {y_1}} \right) + c = 0$$
Proof: Using Calculus method
Equation of circle is ${x^2} + {y^2} + 2gx + 2fy + c = 0...(1)$
Since point $P({x_1},{y_1})$ lies on the circle, it satisfies the equation of circle $${x_1}^2 + {y_1}^2 + 2g{x_1} + 2f{y_1} + c = 0...(2)$$
Differentiate equation $(1)$ with respect to $x$, we get $$2x + 2y\frac{{dy}}{{dx}} + 2g + 2f\frac{{dy}}{{dx}} + 0 = 0$$ $$\frac{{dy}}{{dx}} = - \left( {\frac{{x + g}}{{y + f}}} \right)$$ $${\frac{{dy}}{{dx}}_{\left( {{x_1},{y_1}} \right)}} = - \left( {\frac{{{x_1} + g}}{{{y_1} + f}}} \right)...(3)$$
Therefore, equation of line at a point $P({x_1},{y_1})$, whose slope is known can be written as $$y - {y_1} = {\frac{{dy}}{{dx}}_{\left( {{x_1},{y_1}} \right)}}\left( {x - {x_1}} \right)$$
Put value of slope from equation $(3)$
$$y - {y_1} = - \left( {\frac{{{x_1} + g}}{{{y_1} + f}}} \right)\left( {x - {x_1}} \right)$$ $$\left( {y - {y_1}} \right)\left( {{y_1} + f} \right) + (x - {x_1})({x_1} + g) = 0$$ $$x{x_1} + y{y_1} + gx + fy = {x_1}^2 + {y_1}^2 + g{x_1} + f{y_1}$$
Add $g{x_1} + f{y_1} + c$ on both sides, we get
$$x{x_1} + y{y_1} + gx + fy + g{x_1} + f{y_1} + c{\text{}} = {x_1}^2 + {y_1}^2 + 2g{x_1} + 2f{y_1} + c$$ $$x{x_1} + y{y_1} + gx + fy + g{x_1} + f{y_1} + c{\text{}} = 0$$ $$x{x_1} + y{y_1} + g\left( {x + {x_1}} \right) + f\left( {y + {y_1}} \right) + c = 0$$
NOTE: This method of finding equation of tangent at any point is applied only for any conics of second degree i.e., equation of tangent of $a{x^2} + b{y^2} + 2gx + 2fy + 2hxy + c = 0$ at $({x_1},{y_1})$ is $$ax{x_1} + by{y_1} + g\left( {x + {x_1}} \right) + f\left( {y + {y_1}} \right) + h\left( {x{y_1} + y{x_1}} \right) + c = 0$$
Question 16. Prove that the tangent to the circle ${x^2} + {y^2} = 5$ at the point $(1,-2)$ also touches the circle ${x^2} + {y^2} - 8x + 6y + 20 = 0$ and also find the point of contact.
Solution: The equation of tangent at point $(1,-2)$ to the circle ${x^2} + {y^2} = 5$ using point form is $$x - 2y = 5$$
Now, to prove that it is also the tangent to the circle ${x^2} + {y^2} - 8x + 6y + 20 = 0$, centre of circle is $(4,-3)$ and 
radius $r = \sqrt {{g^2} + {f^2} - c} = \sqrt { - {4^2} + {3^2} - 20} = \sqrt 5 $.
radius $r = \sqrt {{g^2} + {f^2} - c} = \sqrt { - {4^2} + {3^2} - 20} = \sqrt 5 $.If the line $x-2y=5$ is tangent to the circle then the perpendicular distance form centre $(4,-3)$ to the line is equal to the radius of circle.
Therefore, the perpendicular distance form centre $(4,-3)$ to the line$ = \left| {\frac{{4 + 6 - 5}}{{\sqrt 5 }}} \right| = \frac{5}{{\sqrt 5 }} = \sqrt 5 $, which is equal to the radius of circle.
Now, let us assume the tangent $x-2y=5$ touches the circle ${x^2} + {y^2} - 8x + 6y + 20 = 0$ at point $B(\alpha ,\beta )$, which satisfies both the equation of tangent and the circle.
$$\alpha - 2\beta = 5...(1)$$ $${\alpha ^2} + {\beta ^2} - 8\alpha + 6\beta + 20 = 0...(2)$$
Put value of $\alpha $ from equation $(1)$ in equation $(2)$, we get
$${\left( {5 + 2\beta } \right)^2} + {\beta ^2} - 8\left( {5 + 2\beta } \right) + 6\beta + 20 = 0$$ $${\beta ^2} + 2\beta + 1 = 0$$ $${\left( {\beta + 1} \right)^2} = 0$$
Therefore, $\beta = - 1$
Put the value of $\beta $ in equation $(1)$, we get $\alpha = 3$.
So, the point of contact is $(3,-1)$.
Question 17. Find the equation of tangent to circle ${x^2} + {y^2} - 80x - 60y + 2100 = 0$ at the point nearest to the origin.
Solution: From the equation of circle given, the centre of circle is $(40,30)$ and radius
$$r = \sqrt {{g^2} + {f^2} - c} = \sqrt { - {{40}^2} + - {{30}^2} - 2100} = \sqrt {400} = 20$$
Distance of origin $(0,0)$ from centre of circle $(40,30)$$ = OC = \sqrt {{{(40 - 0)}^2} + {{(30 - 0)}^2}} = \sqrt {1600 + 900} = \sqrt {2500} = 50$
As shown in figure $27$, $OR = OC - RC = 50 - 20 = 30$
From figure $28$, to find the coordinates of point $R({x_1},{y_1})$ apply section formula between points $O$ and $C$, we get
$${x_1} = \frac{{\left( {40 \times 3} \right) + 0}}{5} = 24$$ and $${y_1} = \frac{{90 + 0}}{5} = 18$$
Therefore, the coordinates of point $R({x_1},{y_1})$ nearest to the origin is $(24,18)$.
Now we use $T=0$ to find the equation of tangent at point $R(24,18)$ i.e.,
$$24x + 18y - 40\left( {x + 24} \right) - 30\left( {y + 18} \right) + 2100 = 0$$
The equation of tangent is $$4x + 3y = 150$$
