Find the equations of the lines that pass through the point $(1,3)$ and are tangent to the circle $x^{2}+y^{2}=2$

Question

Since the line passes through $(1,3)$ I substituted: $3=m+b$ so $m=3-b$ and $y=(3-b)x+b$. But if I then plug the line equation into the circle equation and take the discriminant, I end up with terms in the 4th power, which doesn't help me solve the problem.

Answer 1

A straight line with slope $m$ through the point $(1,3)$ has equation: $$y = m(x-1)+3$$ This line is tangent to the circle $x^2+y^2=2$ if the equation for the abscissae of intersection points has a double root. This equation is the quadratic equation: $$x^2+(mx-m+3)^2=2\iff(m^2+1)x^2-2m(m-3)x+m^2-6m+7=0. $$ It has a double root if and only if its (reduced) discriminant is $0$: $$\Delta'=m^2(m-3)^2-(m^2+1)(m^2-6m+7)=m^2+6m-7.$$ $1$ is clearly a root of $\Delta'$, hence the other root is $-7$. The double root of the equation in $x$ is then $$x=\dfrac{m(m-3)}{m^2+1}=\begin{cases}-1&\text{if} \enspace m=1,\\\dfrac75&\text{if} \enspace m=-7\end{cases}. $$ Finally the ordinate is calculated with the equation of the line and finally obtain as points of contact: $$(-1,1)\enspace\text{and}\enspace \biggl(\dfrac75,\dfrac15\biggr).$$

Answer 2

Assume the slope of the line to be m.

The concerned line is $$y-3=m(x-1)$$ $$mx-y+3-m=0$$

The distance from center to this line would be equal to r where r is radius of circle.

$$ \frac{|3-m|}{\sqrt{1+m^2}}=\sqrt2 $$ Solve this equation and calculate values of m and substitute in above equation.

Answer 3

Hint:

The discriminant of the system of the equations of a straight line containing one parameter ( a bundle of straight lines) and a circumference is always an equation of second degree in the parameter, because there can be at most two tangent to a circumference passing through a point. So, if you find an equation of degree $4$ (or $3$) this means that you have an error in your calculations....Redo, the terms of degree $>2$ must disappear !

Answer 4

Consider tangent point (x0,y0)

Step1) The equation of tangent is xx0+yy0=2

Step2) the tangent pass (1,3) so x0+3y0=2

Step3) the tangent point lies on the circle so: x0^2+y0^2=2

Step4) solve the system x0+3y0=2 and x0^2+y0^2=2

    Giving (x0,y0)= (−1,1)and(7/5,1/5)