# Tangent to a Circle

## Condition of Tangency of a Straight Line to 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. Now, the straight line will touch the circle (the line will be a tangent to the circle) if it intersects the circle in two coincident points i.e. if the roots of the above quadratic equation are same.

$\text{i.e. if }B^2-4AC=0$ $4m^2c^2-4(1+m^2)(c^2-a^2)=0$ $m^2c^2-(c^2-a^2+m^2c^2-a^2m^2)=0$ $m^2c^2-c^2+a^2-m^2c^2+a^2m^2=0$ $c^2=a^2(1+m^2)$ $\therefore c=\pm a\sqrt{1+m^2}$ This is the condition of tangency. And, the equation of the tangent line is, $y=mx\pm a\sqrt{1+m^2}$ This equation of the tangent is in slope form.

## Equation of the Tangent to the Circle $x^2+y^2=a^2$ at a Point $(x_1,y_1)$ on the Circle

Let $P(x_1,y_1)$ be a point on the circle $x^2+y^2=a^2$ Differentiating with respect to $x$, $2x+2y\frac{dy}{dx}=0$ $y\frac{dy}{dx}=-x$ $\therefore\frac{dy}{dx}=-\frac{x}{y}$ Since the point $P(x_1,y_1)$ lies on the circle, $\text{Slope of tangent }(m)=\left|\frac{dy}{dx}\right|_{(x_1,y_1)}=-\frac{x_1}{y_1}$ Hence, the equation of the tangent line is, $y-y_1=m(x-x_1)$ $y-y_1=-\frac{x_1}{y_1}(x-x_1)$ $yy_1-y_1^2=-xx_1+x_1^2$ $xx_1+yy_1=x_1^2+y_1^2$ $\therefore xx_1+yy_1=a^2$ This is the required equation of the tangent in point form.

## Equation of Tangent to the General Circle $x^2+y^2+2gx+2fy+c=0$ at a Point $(x_1,y_1)$ on the Circle

Let $P(x_1,y_1)$ be a point on the circle $x^2+y^2+2gx+2fy+c=0$ Differentiating with respect to $x$, $2x+2y\frac{dy}{dx}+2g+2f\frac{dy}{dx}+c=0$ $(y+f)\frac{dy}{dx}=-(x+g)$ $\frac{dy}{dx}=-\frac{x+g}{y+f}$ Since the point $P(x_1,y_1)$ lies on the circle, $\text{Slope of tangent }(m)=\left|\frac{dy}{dx}\right|_{(x_1,y_1)}=-\frac{x_1+g}{y_1+f}$

Hence, the equation of the tangent line is, $y-y_1=m(x-x_1)$ $\text{or, }y-y_1=-\frac{x_1+g}{y_1+f}(x-x_1)$ $\text{or, }(y-y_1)(y_1+f)=-(x_1+g)(x-x_1)$ $\text{or, }yy_1+fy-y_1^2-fy_1=-xx_1+x_1^2-gx+gx_1$ $\text{or, }xx_1+yy_1+gx+fy=x_1^2+y_1^2+gx_1+fy_1$ $\text{or, }xx_1+yy_1+gx+gx_1+fy+fy_1+c$ $=x_1^2+y_1^2+2gx_1+2fy_1+c$ $\therefore xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0$ This is the required equation of the tangent in point form.

## Length of the Tangent from an External Point to a Circle

Let a tangent $PT$ be drawn from an external point $P(x_1,y_1)$ to the circle $x^2+y^2+2gx+2fy+c=0$ The centre of the circle is $C(-g,-f)$ and its radius $=\sqrt{g^2+f^2-c}$. $\therefore PC^2=(x_1+g)^2+(y_1+f)^2$ $\therefore CT^2=g^2+f^2-c^2$ $\text{Also }CT\perp PT,$ $\therefore PT^2=PC^2-CT^2$ $=(x_1+g)^2+(y_1+f)^2-(g^2+f^2-c)$ $=x_1^2+2gx_1+y_1^2+2fy_1+c$ $=x_1^2+y_1^2+2gx_1+2fy_1+c$ Hence the length of the tangent $PT$ is $\sqrt{x_1^2+y_1^2+2gx_1+2fy_1+c}$ If the equation of the circle be $x^2+y^2=a^2$, then the length of the tangent from an external point $(x_1,y_1)$ can be similarly shown to be equal to $\sqrt{x_1^2+y_1^2-a^2}$

### Prove that the line $5x+12y+78=0$ is tangent to the circle $x^2+y^2=36$.

Given equation of circle is $x^2+y^2=36$ $\therefore \text{Centre }C(h,k)=(0,0)$ $\text{Radius }(r)=6$ and, given line is, $5x+12y+78=0$ Now, perpendicular distance of this line from $C(0,0)$ is $d=\pm\left(\frac{5×0+12×0+78}{\sqrt{25+144}}\right)=\pm\frac{78}{3}=\pm 6$ $\therefore d=\pm r$ Hence, the line is tangent to the circle.

### Find the equation of tangent to the circle $x^2+y^2=36$ at $(-6,0)$.

Given point $P(x_1,y_1)=(-6,0)$. And, given circle is $x^2+y^2=36$ Hence the equation of tangent is $xx_1+yy_1=a^2$ $-6x+0=36$ $-6x=36$ $\therefore x+6=0$

### Find the equation of tangent to the circle $x^2+y^2-3x+10y-15=0$ at $(4,-11)$.

Given point $P(x_1,y_1)=(4,-11)$. And, given circle is $x^2+y^2-3x+10y-15=0$ $\begin{array}{c}\therefore 2g=-3, & 2f=10 &\text{and}& c=-15 \\ g=-\frac{3}{2}, & f=5 &\text{and}& c=-15\end{array}$ Hence the equation of tangent is $xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0$ $4x–11y-\frac{3}{2}(x+4)+5(y-11)-15=0$ $8x-22y-3x-12+10y-110-30=0$ $\therefore 5x-12y-152=0$

### Find the equation of the tangents to the circle $x^2+y^2=4$, which are parallel to $3x+4y-5=0$.

Here, given line is, $3x+4y-5=0$ $\therefore\text{Slope of given line}=-\frac{3}{4}$ Since the tangent is parallel to the given line. $\therefore\text{Slope of tangent }(m)=-\frac{3}{4}$ The equation of tangent is given by $y=mx+c$ $y=-\frac{3}{4}x+c$ $3x+4y-4c=0\text{ __(1)}$ Given circle is, $x^2+y^2=4$ $\therefore \text{Centre }C(h,k)=(0,0)$ $\text{and, Radius }(r)=2$ Since line $\text{(1)}$ is tangent to the circle, so $r=\pm\left(\frac{3×0+4×0-4c}{\sqrt{9+16}}\right)$ $2=\pm\frac{4c}{5}$ $\therefore c=\pm\frac{5}{2}$ Hence, the tangents are $3x+4y\pm 4×\frac{5}{2}=0$ $\therefore 3x+4y\pm 10=0$

### Determine the length of the tangents to the circle $x^2+y^2+4x+6y-19=0$ from $(6,4)$.

Given circle is $x^2+y^2+4x+6y-19=0$ from point $P(x_1,y_1)=(6,4)$ Hence, length of the tangent from $(6,4)$ is $\sqrt{x_1^2+y_1^2+4x_1+6y_1-19}$$=\sqrt{36+16+24+24-18}=9$