Tangent to a Circle

Condition of Tangency of a Straight Line 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

Equation of Tangent to a 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

Length of 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 \[\sqrt{x_1^2+y_1^2+4x_1+6y_1-19}\]\[=\sqrt{36+16+24+24-18}=9\]


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