Find the equation of the circle with radius $4$ units, whose centre lies on the line $4x+13y=32$ and which touches the line $4x+3y+28=0$.
My Approach:
Radius $r=4$ units
Let $P(h,k)$ be the centre of the circle. Then $4h+13k=32$.
Please help me to move further.
The equation of a circle with center in $(a,b)$ and radius $4$ is
$$
\left( {x - a} \right)^{\,2} + \left( {y - b} \right)^{\,2} = r^{\,2} = 16
$$
Now you must have
$$
\left\{ \begin{gathered}
4a + 13b = 32\quad \text{(center}\,\text{on}\,\text{the}\,\text{line}\,(\text{a))} \hfill \\
4x + 3y + 28 = 0\quad \text{(point}\,\text{on}\,\text{the}\,\text{line}\;\text{(b))} \hfill \\
\left( {x - a} \right)^{\,2} + \left( {y - b} \right)^{\,2} = 16\quad \text{(point}\,\text{on}\,\text{the}\,\text{circle)} \hfill \\
\end{gathered} \right.
$$
Now, from the first express e.g. $a$ in function of $b$, and from the second e.g. $y$ in function of $x$ and place them in the third. You will get
$$\left( {x - 8 - \frac{{13}}
{4}b} \right)^{\,2} + \left( { - \frac{{28}}
{3} - \frac{4}
{3}x - b} \right)^{\,2} = 16
$$
Expand as a quadratic equation in $x$ and impose to have two coincident solutions, i.e. find $b$ such that the discriminant be $0$.
You will find two values, corresponding to whether the circle is on one side or the other with respect to the crossing point of the two lines.
Use each value of $b$ to determine the corresponding $x$, $y$, $a$.
From L :4x + 3y + 28 = 0, get $\alpha (= 53.13^0$, but not really necessary). Note further that the blue triangle is a 3-4-5 right triangle.
Find G where L cuts the y-axis. OG is then known.
L’ is normal to L and passes through G. The angle between L’ and the y-axis = $\beta = \alpha$.
A point on the red line will be the center of the required circle. It will cut the y-axis at T, whose co-ordinates are to be determined.
The blue and green triangles are similar. Therefore, TG is known (using the 3-4-5 properties).
Find OT and therefore $T = … = (0, \dfrac {-8}{3})$.
Form the equation of Q which passes through T and has the same slope as L.
Solve P and Q to get X which is the center of the required circle.
...
$4x+13y=32\iff \dfrac y4=\dfrac{8-x}{13}=m$(say)
So, any point on $4x+13y=32$ will be $(8-13m,4m)$
Now the perpendicular distance of $(8-13m,4m)$ from $4x+3y+28=0$ will be $$\dfrac{|4(8-13m)+3(4m)+28|}{\sqrt{4^2+3^2}}$$ which needs to be $=$ radius i.e., $4$ unit
Anypoint of the line $4x+13y=32$ have the form $(x,\frac{-31}{100}x+\frac{246}{100})$ and anypoint of the line $4x+3y=-28$ have the form $(x,\frac{-133}{100}x-\frac{933}{100})$. Now need calculate the distance of the point $(x,\frac{-31}{100}x+\frac{246}{100})$ to the point $(x,\frac{-133}{100}x-\frac{933}{100})$, use the euclidean distance equation: $$\sqrt{(x-x)^2+(\frac{-31}{100}x+\frac{246}{100}-(\frac{-31}{100}x+\frac{246}{100}))^2}=4$$ Finding $x$: $$x=-\frac{1579}{102} \lor x=-\frac{779}{102}$$ So exist 2 circles with center in the line $4x+13y=32$ and radius 4 that touch the line $4x+3y=-28$. To find the equations of the circles you need replace the $x$ that we find in the line $4x+13y=32$ for get the center points, and use the equation $(x-h)^2+(y-k)^2=r^2$
