Circle touching tangent

Question

Problem-

$A=(5,5),B=(14,17).$ A circle passes through $A$ and $B$ and touches the line $u:4x+3y-11=0.$ Find the coordinates of the point of contact.

I tried this but could not think of an efficient method to solve this problem. While solving the equations I got this awful system of equations -

$(4x+3y-11)^2=25(x-14)^2+25(y-17)^2$

$(4x+3y-11)^2=25(x-14)^2+25(y-17)^2$

Can someone please give a hint atleast? I am not asking for a full solution. Thanks.

Answer 1

$AB$ is a chord of the circle so the center must lie on the perp bisector of $AB$. Midpoint of $AB$ is $\left(\frac{19}{2}, 11\right)$ and the slope of a line perp to $AB$ is $- \dfrac{3}{4}$.

Now find the equation of the perp bisector in the form $y = - \frac 34x + c$ and then coordinates of the center $O$ is $\left (h, - \frac {3h}4 + c \right)$.

If the foot of the perp from $O$ to the tangent line $4x + 3y - 11 = 0$ is $H$ then $|OH| = r$, where $r$ is the radius of the circle.

That gives you $r$ in terms of $h$. Now given $|OA| = r$

$(h - 5)^2 + (- \frac{3h}{4} + c - 5)^2 = r^2$

That should give you $h$ and from there on, you can easily find coordinates of $O$. Finally, coordinates of $H$ is the intersection of the given tangent line and the line perp to it through $O$.

Alternate approach:

Find intersection of the line through $AB$ and the given tangent line. Say it is $X$. Then using power of the point $X$, $XA \cdot XB = XH^2$ where $H$ is the point of contact you need to find coordinates of.

Given you now know $XH$, coordinates of $X$ and equation of the line through $XH$, you can easily find coordinates of $H$.

Answer 2

The equation of a circle (with centre $(a,b)$ and radius $r$) is $$(x-a)^2+(y-b)^2=r^2$$ for some $a,b,r\in\mathbb{R},r>0$.

We need to find three values, $a,b,r$, so we'll need three pieces of information (i.e. three equations) to do so.

• The first piece of information we have is that $x=5,y=5$ satisfies the equation (i.e. $A$ lies on the circle).
• The second piece of information we have is that $x=14,y=17$ satisfies the equation.
• The third piece of information is that there is exactly one solution to the simultaneous equations $(x-a)^2+(y-b)^2=r^2$ and $4x+3y-11=0$ i.e. the circle intersects with $u$ at exactly one point (so the line $u$ is a tangent). You can rearrange the equation of $u$ to get an equation of the form $y=...$ and then substitute into the circle equation to get a quadratic in $x$. When does a quadratic in $x$ have exactly one solution?

Once you have simplified these three equations as far as possible, separately, consider carefully what the best way to use elimination or substitution to find the solution set for $a,b,r$ is.