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irina [24]
3 years ago
12

Find the equation of the normal to the circle, whose equation is given below, at the point (1,2).( x − 3 ) 2 + ( y + 2 ) 2 = 20

Mathematics
1 answer:
grandymaker [24]3 years ago
8 0

Answer:

The equation of the normal is y = 4 - 2·x

Step-by-step explanation:

The given equation of the circle, is presented as follows;

(x - 3)² + (y + 2)² = 20

The point of the normal of the circle = (1, 2)

The equation of the normal to a circle, x² + y² + D·x + E·y + F = 0 at a point P(x₁, y₁) is given as follows;

\dfrac{y - y_1}{x - x_1} = \dfrac{2 \cdot y_1 + D}{2 \cdot x_1 + E}

Expanding the given equation of the circle, gives;

(x - 3)² + (y + 2)² = x² + y² - 6·x + 4·y + 13 = 20

∴ x² + y² - 6·x + 4·y + 13 - 20 = x² + y² - 6·x + 4·y - 7 = 0

x² + y² - 6·x + 4·y - 7 = 0

∴ x₁ = 1, y₁ = 2, D = -6, E = 4, and F = 13

Which gives;

\dfrac{y - 2}{x - 1} = \dfrac{2 \times 2 + 4}{2 \times  1 + (-6)} = \dfrac{8}{-4} = -2

∴ y - 2 = -2 × (x - 1) = 2 - 2·x

y = 2 - 2·x + 2 = 4 - 2·x

The equation of the normal to the circle with equation (x - 3)² + (y + 2)² = 20, at the point (1, 2) is y = 4 - 2·x

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Answer:

The perimeter of the triangle ABC is 17 cm.

Step-by-step explanation:

Consider the Isosceles triangle ABC.

The sides CA and CB are equal with measures, 5 cm.

The base angles are assumed to be <em>x</em>° each. Hence, the angle ACB is 2<em>x</em>°.

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Consider the right angled triangle ACP.

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<em>x</em>° + <em>x</em>° + 90° = 180°

2<em>x</em>° = 90°

<em>x</em>° = 45°

Compute the length of side AP as follows:

cos\ 45^{0}=\frac{AP}{CA}

      \frac{1}{\sqrt{2}}=\frac{AP}{5}

     AP =\frac{5}{\sqrt{2}}\\\\AP=3.5

Then the length of side AB is:

AB = AP + PB

     = 3.5 + 3.5

     = 7 cm

The perimeter of triangle ABC is:

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     The number "x" is given by:
y= \left \{ {{y_1=25-13 \rightarrow~y_1=12} \atop {y_1=25-12 \rightarrow ~y_1=13}} \right.
  
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\boxed {S=({12,13})} \rightarrow~The~curly~braces~does~not~want~appear~here.
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