Question

In a circle if 10 c.m diameter is drawn horizontally and 8c.m diameter is drawn vertically it intersects.( Diameter doesn't go through the centre of the circle)
What's the radius of the circle?

Answer 1

Answer on Question #80596 - Math - Geometry

Question

In a circle if $10 \, \text{cm}$ diameter is drawn horizontally and $8 \, \text{cm}$ diameter is drawn vertically it intersects. (Diameter doesn't go through the centre of the circle) What's the radius of the circle?

Solution

AB = 10, CD = 8, AO = r = ?

The triangles of the AHO and $\mathsf{DH}^{\prime}\mathsf{O}$ are similar.

AH/OH' = HO/H'D

$\mathrm{OH}^{\prime} = \mathrm{y}, \mathrm{HO} = \mathrm{x}, 5 / \mathrm{y} = \mathrm{x} / 4, \mathrm{y} = 20 / \mathrm{x}$

$$\left\{ \begin{array}{l} x ^ {2} + 2 5 = r ^ {2} \\ y ^ {2} + 1 6 = r ^ {2} \end{array} \right.$$ $$x ^ {\wedge} 2 - y ^ {\wedge} 2 + 9 = 0$$ $$x ^ {\wedge} 2 - 4 0 0 / x ^ {\wedge} 2 + 9 = 0$$ $$x ^ {\wedge} 4 + 9 ^ {*} x ^ {\wedge} 2 - 4 0 0 = 0$$ $$x ^ {\wedge} 2 = t, t \geq 0$$ $$t ^ {\wedge} 2 + 9 ^ {*} t - 4 0 0 = 0$$ $$D = b ^ {\wedge} 2 - 4 ^ {*} a ^ {*} c$$ $$t 1 = (- b + (D) ^ {\wedge} 0. 5) / 2 ^ {*} a$$ $$t 2 = (- b - (D) ^ {\wedge} 0. 5) / 2 ^ {*} a$$

D = 1681

t1 = 16

t2 = -25

$$\left\{ \begin{array}{c} t1 = 16, t = -25 \\ t \geq 0 \end{array} \right., t = 16$$

16 = x^2

x = ±4 (We take a positive value, because there are no negative lengths in geometry)

y = 20/4 = 5

r = AO = ((AH)^2 + (HO)^2)^0.5 = (16 + 25)^0.5 = 6.4.

Answer: r = 6.4.

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