Find the equation of circle touching given lines and a given point.

Question

$U: 3x+4y-20=0$ and $v:3x+4y+10=0$ are two straight lines. Find the equation of circles touching the given lines and passing through point $P(1,2)$.

Answer 1

Since the lines are parallel, the center of circle lays between these two, on the line $3x+4y+\frac{-20+10}{2}=0$. It's radius is the distance form $3x+4y-5=0$ to one of that lines: $$\frac{|3x+4y-20|}{\sqrt{3^2+4^2}}=\frac{|5-20|}{5}=3$$ So we find the center $(a,b)$:\begin{cases}3a+4b-5=0\\(a-1)^2+(b-2)^2=3^3\end{cases} Can you take it from here?

Answer 2

these are two parallel lines. first we will find a circle touching the two lines without the circle going through the point $(1,2).$ we may be able translate the circle to make it go through the point $(1,2).$

the points on contact is a diameter of the circle and goes through the origin. contact points are of the form $(3t, 4t).$ subbing in $3x+4y = 20 \to t = 4/5$ and $3x+4y = -10 \to t = -2/5.$ therefore the center is $(3/5, 4/5)$ and the radius is $3.$

$$(x-3/5)^2 + (y-4/5)^2 = 9$$ touches the line $3x+4y = 20$ at $(12/5, 16/5)$ and the line $3x+4y=-10$ at $(-6/5, -8/5).$

now we will move the center parallel to the lines $3x+4y = 0$ by making the center $(3/5+4k/5, 4/5 -3k/5).$ the equation of the circle is $$(x - (3+4k)/5)^2 +(y-(4-3k)/5)^2 = 9.\tag 1 $$ we will fix $k$ by making $(1)$ go through $(1,2).$ that is $$(2-4k)^2+(1+3k)^2=225 \to k = \frac{1 \pm \sqrt{221} }5.$$ so there are two circles $(1) $ corresponding to $k = \frac{1 \pm \sqrt{221} }5.$