The line with the equation $2x-3y=10$ touches the circle with center $M(-2,4)$ at point $A$. Find the equation of $A$.

Question

This is one of the last problems on my sheet but i can't seem to understand it. Several of my friends have tried explaining it to me but i still can't seem to get it, i think i need some new perspectives so i could better understand this problem ://

Answer 1

The essential point is that the line from the center of a circle to the point of contact of a tangent is perpendicular to the tangent. Call this line L. If we can find the equation of L, then we know that the point A lies on both L and the line $2x-3y=10$ so solving these equation simultaneously will give A.

We know the tangent has slope $\frac23$ so L has slope $-\frac32.$ The equation of L is of the form $y=-\frac32 x + b$. Since we know that $(-2,4)$ lies on L, we can solve for $b$ to get the equation of L, then we can solve for A as described above.

Answer 2

Option:

1)$ (x+2)^2+ (y-4)^2=r^2.$

Differentiate with respect to $x:$

$2(x+2)+2(y-4)dy/dx =0$.

At point of tangency $A$:

$dy/dx= 2/3$ , slope of $2x-3y=10$.

Hence:

$2(x+2) +2(y-4)(2/3) =0$, or

$6(x+2) +4(y-4)=0$,

$6x+4y= 4.$

2)Tangent line : $2x-3y=10.$

Solve the simultaneous equations to find $A(x,y).$

Note:

Implicit differentiation yields:

$2(x+2)+2(y-4)m= 0$, where $dy/dx=m$, slope of the given line.

Hence

$(1/m)(x+2) +(y-4)=0, $

a straight line passing through the centre of the circle $M(-2,4)$ with slope $-(1/m)$.

Answer 3

The method given by saulspatz is definitely easier!

Here is an alternative method: the distance from the point $M(-2,4)$ to the tangent line $2x - 3y = 10$ is: $$r=\frac{|2(-2)-3(4)-10|}{\sqrt{2^2+(-3)^2}}=\frac{26}{\sqrt{13}}=2\sqrt{13}.$$ The equation of a circle with center $M(-2,4)$ and the radius $2\sqrt{13}$ is: $$(x+2)^2+(y-4)^2=52.$$ The tangent point $A$ is: $$\begin{cases}2x - 3y = 10\\(x+2)^2+(y-4)^2=52\end{cases} \Rightarrow A(2,-2).$$