1)Find equation of circle
2)Equation of another tangent from point $H$ to the circle
The circle in the first quadrant touches $x$-axis $y$-axis and straight line $3x-4y-20=0$. The point $H(12,4)$ lies on the straight line. How should I proceed to get the equation?
$y=(3/4)x-5$. I know $c=-5$. Correct me if I am wrong?
On
1) We want to find the $x-$ and $y-$intercepts of the line so that we can use geometry to solve for the radius and center. These are simply $(0, -5)$ and $(\frac{20}{3}, 0).$ The hypotenuse of the right triangle formed is $\frac{25}{3}$ by Pythagoras. The circle divides each side of the triangle into two parts, and every pair of sides shares a segment. Call the radius $r$ and the other segments $a$ and $b.$ Then we have $r + a = \frac{20}{3},$ $r + b = 5,$ and $a + b = \frac{25}{3}.$ Solving, we have $a + b + r = 10,$ so $r = 10 - \frac{25}{3} = \frac{5}{3}.$ This means the center is at $(\frac{5}{3},-\frac{5}{3}).$ This is actually in Quadrant IV - it is impossible for the circle to be in Quadrant I. The equation of the circle is thus $\boxed{(x - \frac{5}{3})^{2} + (y + \frac{5}{3})^{2} = \frac{25}{9}}.$
On
Since the circle touches x-axis and y-axis its center must lie on line $x=y$. Its coordinates must be $ (h,h)$. Equation of circle is
$$ (x-h)^2 +(y-h)^2 = h^2 $$
It should satisfy line L $$ y = 3 x/4 -5 $$
Eliminate y:
$$ x= 4/25 (15+7 h+2 \sqrt 2 \sqrt{-50-5 h+3 h^2}) $$
The quantity under radical sign should vanish for a coincident root at tangent point, or ,
$$ -50-5 h+3 h^2 = 0 $$
which has roots:
$$ h = 5 , h = -10/3 $$
The latter is discarded as the circle must be in first quadrant.
So the equation of the desired circle is finally:
$$ (x-5)^2 +(y-5)^2 = 5^2 = 25 $$
Point of tangential contact $(8,1)$ (not asked for).
For question 1) we look for circles of the form $(x-a)^2+(y-a)^2=a^2$, since these are tangent to the axes, and the only ones because a tangency to the axis determines the radius line.
($(x-a)^2+(y+a)^2=a^2$ gives the circles in the 2. and 4. quadrants)
The problem is to get the third tangency with the given line. First the point needs to be on the circle: $(x-a)^2+((3/4)x-5-a)^2=a^2$ or $(\frac{25}{16}x^2+(-\frac{7}{2}a-\frac{15}{2})x+a^2+10a+25=0$, and it should touch doubly, i.e. the discriminant should be 0. This gives us a 2. order equation in a. In fact it is $(-\frac{7}{2}a-\frac{15}{2})^2-4(\frac{25}{16})(a^2+10a+25)=2 (a - 5) (3 a + 10)=0$, so that $a=5$ or $a=-10/3$ a solution in the third quadrant.
So the solution is $(x-5)^2+(y-5)^2=5^2$ which goes through $(8,1)$ on the line which is a $3,4,5$-triangle from the center $(5,5)$.
For question 2) look at lines $y-4=k(x-12)$ and substitute into $(x-5)^2+(y-5)^2=5^2$, i.e. $(x-5)^2+(4+k(x-12)-5)^2=5^2$ which simplifies to a quadratic in $x$ with discriminant $-12k^2-7k+12$ which has solutions $k=-\frac{4}{3}$ and $k=\frac{3}{4}$. Putting it all together we get $3y+4x-60=0$ and the original line.