# A Point, a Line and a Circle

## A Point and a Circle

Let the equation of a circle be $x^2+y^2+2gx+2fy+c=0$ Hence, the centre of the circle is $C(-g,-f)$ and radius $r=\sqrt{g^2+f^2-c}$ Let $P(x_1,y_1)$ be any point. Denoting $x^2+y^2+2gx+2fy+c=0$ by $F(x,y)$, $F(x_1,y_1)=x_1^2+y_1^2+2gx_1+2fy_1+c$ $=(x_1+g)^2+(y_1+f)^2-(g^2+f^2-c)$ $=CP^2-r^2$ Therefore,

• If $CP^2=r^2$ $[F(x_1,y_1)=0]$, then $P$ lies on the circle.
• If $CP^2<r^2$ $[F(x_1,y_1)<0]$, then $P$ lies inside the circle.
• If $CP^2>r^2$ $[F(x_1,y_1)>0]$, then $P$ lies outside the circle.

For a circle $x^2+y^2=a^2$, any point $(x_1,y_1)$ will lie inside, on or outside of the circle according as $x_1^2+y_1^2$ is $<$ $=$ or $>$ $a^2$.

## A Line and a Circle

Let the equation of a circle be $x^2+y^2=a^2$ and, the equation of a line be $y=mx+c$ Now, let us solve these two equations simultaneously to obtain the points of intersection of the circle and the line. For this, eliminating $y$, $x^2+(mx+c)^2=a^2$ $x^2+m^2x^2+2mcx+c^2-a^2=0$ $(1+m^2)x^2+2mcx+(c^2-a^2)=0\text{ __(1)}$ This is quadratic in $x$. Hence, $x$ has two values and corresponding to the two values of $x$, we can obtain the two values of $y$. In this way, we can obtain the points of intersection.

Now, the determinant of equation $\text{(1)}$ is $B^2-4AC$ $=4m^2c^2-4(1+m^2)(c^2-a^2)$ $=4(m^2c^2-c^2+a^2-m^2c^2+m^2a^2)$ $=4[a^2(1+m^2)-c^2]$

• If $a^2(1+m^2)>c^2$, then the roots of $\text{(1)}$ i.e. the values of $x$ are real and distinct. In this case, the line cuts the circle at two distinct points.
• If $a^2(1+m^2)=c^2$, then the roots of $\text{(1)}$ are real and equal. In this case the line meets the circle at two coincident points. We say that the line is tangent to the circle.
• If $a^2(1+m^2)<c^2$, then the roots of $\text{(1)}$ are imaginary. In this case, the line does not intersect the circle. But it is customary to say that the line cuts the circle at the imaginary points.

Hence, the line will touch the circle if it intersects the circle in two coincident points i.e. if the roots are real and equal. $\therefore B^2-4AC=0$ $4[a^2(1+m^2)-c^2=0$ $a^2(1+m^2)-c^2=0$ $c^2=a^2(1+m^2)$ $\therefore c=\pm a\sqrt{1+m^2}$ This is the condition of tangency. Thus, the equation of a line tangent to the circle is given by, $y=mx\pm a\sqrt{1+m^2}$ This equation of the tangent is in slope form. To deduce the equation of the tangent in point form, see this: Equation of The Tangent to the Circle.