Question

consider a triangle ABC with right angle at B with AB=3 and BC=4. A circle 'S' touching side BC is drawn intersecting sides AB at points D and E and the side AC at points F and G respectively.Also D and F are midpoints of sides AB and AC respectively.FIND RADIUS OF CIRCLE 'S',LENGHT OF CHORD EF OF CIRCLE 'S' AND LENGTH OF PORTION FG ON HYPOTENUSE AC IS

Answer 1

Radius = 13/12, Length of chord EF = 13/6, Length of portion FG = 11/10

Answer 2

Let B be the origin (0,0). Since the triangle ABC is right-angled at B, we can place BC along the x-axis and AB along the y-axis.

The coordinates are A=(0,3), B=(0,0), C=(4,0). The length of AC is $\sqrt{3^2 + 4^2} = 5$. The equation of line AB is $x=0$. The equation of line BC is $y=0$. The equation of line AC passing through A(0,3) and C(4,0) is $\frac{x}{4} + \frac{y}{3} = 1$, or $3x+4y-12=0$.

The circle 'S' touches side BC (y=0). Let the center of the circle be (h, k) and the radius be r. Since it touches y=0, the distance from the center to y=0 is the radius, so $|k|=r$. Assuming the circle is in the first quadrant, $k=r$. The center is (h, r). The equation of the circle is $(x-h)^2 + (y-r)^2 = r^2$.

D is the midpoint of AB. A=(0,3), B=(0,0). D = $(\frac{0+0}{2}, \frac{3+0}{2}) = (0, 3/2)$. F is the midpoint of AC. A=(0,3), C=(4,0). F = $(\frac{0+4}{2}, \frac{3+0}{2}) = (2, 3/2)$.

The circle passes through D=(0, 3/2) and F=(2, 3/2). Since D and F have the same y-coordinate, the line segment DF is horizontal. The center of the circle must lie on the perpendicular bisector of the chord DF. The midpoint of DF is $(\frac{0+2}{2}, \frac{3/2+3/2}{2}) = (1, 3/2)$. The length of DF is $2-0=2$. The perpendicular bisector of DF is the vertical line $x=1$. Since the center of the circle is (h, r), we must have $h=1$.

Now the center is (1, r). The circle passes through D=(0, 3/2). $(0-1)^2 + (3/2-r)^2 = r^2$ $(-1)^2 + (3/2)^2 - 2(3/2)r + r^2 = r^2$ $1 + 9/4 - 3r + r^2 = r^2$ $1 + 9/4 = 3r$ $13/4 = 3r$ $r = 13/12$.

The radius of the circle 'S' is $13/12$. The center of the circle is $(1, 13/12)$. The equation of the circle is $(x-1)^2 + (y-13/12)^2 = (13/12)^2$.

Now we find the points E on AB ($x=0$) and G on AC ($3x+4y-12=0$). For points on AB, substitute $x=0$ into the circle equation: $(0-1)^2 + (y-13/12)^2 = (13/12)^2$ $1 + (y-13/12)^2 = (13/12)^2$ $(y-13/12)^2 = (13/12)^2 - 1 = 169/144 - 144/144 = 25/144$ $y - 13/12 = \pm \sqrt{25/144} = \pm 5/12$ $y = 13/12 \pm 5/12$. The two y-coordinates are $y_1 = 13/12 + 5/12 = 18/12 = 3/2$ and $y_2 = 13/12 - 5/12 = 8/12 = 2/3$. The points on AB are D=(0, 3/2) and E=(0, 2/3). The length of the chord DE is the distance between (0, 3/2) and (0, 2/3). Length of DE = $|3/2 - 2/3| = |9/6 - 4/6| = 5/6$. The problem asks for the length of chord EF. E=(0, 2/3) and F=(2, 3/2). Length of EF = $\sqrt{(2-0)^2 + (3/2-2/3)^2} = \sqrt{2^2 + (5/6)^2} = \sqrt{4 + 25/36} = \sqrt{(144+25)/36} = \sqrt{169/36} = 13/6$.

For points on AC, substitute $y = (12-3x)/4$ into the circle equation: $(x-1)^2 + (\frac{12-3x}{4} - \frac{13}{12})^2 = (\frac{13}{12})^2$ $(x-1)^2 + (\frac{3(12-3x)-13}{12})^2 = (\frac{13}{12})^2$ $(x-1)^2 + (\frac{36-9x-13}{12})^2 = (\frac{13}{12})^2$ $(x-1)^2 + (\frac{23-9x}{12})^2 = (\frac{13}{12})^2$ Multiply by $12^2=144$: $144(x-1)^2 + (23-9x)^2 = 13^2$ $144(x^2-2x+1) + (529 - 414x + 81x^2) = 169$ $144x^2 - 288x + 144 + 529 - 414x + 81x^2 = 169$ $225x^2 - 702x + 673 = 169$ $225x^2 - 702x + 504 = 0$. We know F=(2, 3/2) is one intersection point, so $x=2$ is a root. $25x^2 - 78x + 56 = 0$. If $x=2$ is a root, then $(x-2)$ is a factor. Let the other root be $x_G$. $25x^2 - 78x + 56 = (x-2)(25x - 28) = 25x^2 - 50x - 28x + 56 = 25x^2 - 78x + 56$. The roots are $x=2$ and $x=28/25$. For $x_G = 28/25$, the y-coordinate is $y_G = 3 - 3x_G/4 = 3 - 3(28/25)/4 = 3 - 21/25 = (75-21)/25 = 54/25$. The points on AC are F=(2, 3/2) and G=(28/25, 54/25). The length of the portion FG on hypotenuse AC is the distance between F and G. Length of FG = $\sqrt{(2 - 28/25)^2 + (3/2 - 54/25)^2}$ $2 - 28/25 = 50/25 - 28/25 = 22/25$. $3/2 - 54/25 = 75/50 - 108/50 = -33/50$. Length of FG = $\sqrt{(22/25)^2 + (-33/50)^2} = \sqrt{(44/50)^2 + (-33/50)^2} = \frac{1}{50}\sqrt{44^2 + (-33)^2}$ $44 = 11 \times 4$, $33 = 11 \times 3$. Length of FG = $\frac{1}{50}\sqrt{(11 \times 4)^2 + (11 \times 3)^2} = \frac{1}{50}\sqrt{11^2 (4^2 + 3^2)} = \frac{1}{50}\sqrt{121 (16+9)} = \frac{1}{50}\sqrt{121 \times 25} = \frac{1}{50} \times 11 \times 5 = \frac{55}{50} = 11/10$.

Final results: Radius of circle 'S' = 13/12. Length of chord EF = 13/6. Length of portion FG on hypotenuse AC = 11/10.