You are given a circle centered at the point o and a point a lying outside this circle. From point a, two straight lines are drawn

You are given a circle centered at the point o and a point a lying outside this circle. From point a, two straight lines are drawn, tangent to this circle at points m and n, find the radius of this circle if AO = 50 MN = 48 and it is known that AM.

The solution of the problem:

1. Let us graphically represent a problem where the tangents from point A form equal right-angled triangles with a circle ∆AMO = ∆AНO, in which the base AO = 50 and the height on this base MH = НH = MН: 2 = 48: 2 = 24 are known.

2. For a right-angled triangle:

(ОМ ∙ МА) / AO = МН, the ratio of the products of the legs to the base is equal to the height to the base. Substitute the known values:

(ОМ ∙ МА) / 50 = 24,

MA = (50 ∙ 24) / ОМ,

MA = 1200 / ОМ.

3. In ∆AMO by the Pythagorean theorem:

MA ^ 2 = AO ^ 2 – OM ^ 2,

(1200 / ОМ) ^ 2 = 50 ^ 2 – ОМ ^ 2, we pass ОМ = r – the required radius of the circle,

(1200 / r) ^ 2 = 50 ^ 2 – r ^ 2, {multiply both sides of the equality by r2},

(1200 / r) ^ 2 * r ^ 2 = (50 ^ 2 – r ^ 2) * r ^ 2,

1200 ^ 2 = 502 r ^ 2 – r4,

r4 – 50 ^ 2 r ^ 2 + 1200 ^ 2 = 0, {we have a biquadratic equation, we make the replacement r ^ 2 = R},

R ^ 2 – 50 ^ 2 R + 1200 ^ 2 = 0,

D = 50 ^ 2 – 4 ∙ 1 ∙ 1200 ^ 2 = 6250000 – 5760000 = 490000> 0, √490000 = 700,

R1 = (2500 + 700) / 2 = 1600,

R2 = (2500 – 700) / 2 = 900,

r1 = ± √R1 = ± √1600,

r2 = ± √R2 = ± √900, only positive roots are suitable,

r1 = 40,

r2 = 30.

According to the condition AM <OM, then we choose a large value.

Answer: r = 40.