Vtyjkyuk

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?

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 $(frac203, 0).$ The hypotenuse of the right triangle formed is $frac253$ 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 $r + a = frac203,$ $r + b = 5,$ and $a + b = frac253.$ Solving, we have $a + b + r = 10,$ so $r = 10 - frac253 = frac53.$ This means the center is at $(frac53,-frac53).$ This is actually in Quadrant IV - it is impossible for the circle to be in Quadrant I. The equation of the circle is thus $boxed(x - frac53)^2 + (y + frac53)^2 = frac259.$

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 $(frac2516x^2+(-frac72a-frac152)x+a^2+10a+25=0$, 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 $(-frac72a-frac152)^2-4(frac2516)(a^2+10a+25)=2 (a - 5) (3 a + 10)=0$, 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 $k=-frac43$ and $k=frac34$. Putting it all together we get $3y+4x-60=0$ and the original line.