Tangent to a circle

Question

The positive value of $\lambda$ for which the straight line $4x-3y+\lambda=0$ touches the circle $x^2+y^2-4x+6y-3=0$ is $4/3/5/8$?

My attempt: center is $(2,-3)$. Radius is $\sqrt{4+9+3}=4$. So, distance of the center from the tangent should be equal to radius.$$\implies \frac{8+9+\lambda}{5}=4$$$$\implies 17+\lambda=20$$$$\implies\lambda=3$$ I think this is correct. But when I do it with a different method, I don't get the answer. The equation of the given line can be written as $3y=4x+\lambda$, or, $y=\frac43x+\frac\lambda3$. And the equation of tangent to circle with slope $m$ is $y=mx\pm4\sqrt{1+m^2}$. On comparing the two equations, I get $m=\frac43$. $$\implies4\sqrt{1+(\frac43)^2}=\frac\lambda3$$$$\implies\lambda=20$$ What am I doing wrong?

Answer 1

Your first method was fine. The problem with your second method is that tangent equation you used is for a circle centered at the origin, but your circle’s center is at $(2,-3)$. If you translate that equation accordingly, i.e., use $$y+3=m(x-2)\pm4\sqrt{1+m^2}$$ instead, the two solutions will agree.

Answer 2

The first condition should be $$4=\left|\dfrac{4(2)+(-3)(-3)+\lambda}{\sqrt{3^2+4^2}}\right|$$

$$\implies\lambda+17=\pm20$$

So, the other value of $\lambda=-20-17$

In the second method, we need to replace $y$ with $\dfrac{4x+\lambda}3$ in the given equation of the circle to forma quadratic equation in $x$

whose roots represent the abscissa of intersection with the tangent.

For tangency, the roots must coincide.

The relationship you have used is valid for $$x^2+y^2=4^2$$