Math / Geometry

Question47 Em relação a um referencial ortonormado Oxy considera a circunferência de centro em $A(0,3)$ e que passa em $B(3,-1)$ . 47.1. Seja $r$ a reta que é tangente à circunferência no ponto $B$ . Representa a reta $r$ por uma equação na forma reduzida. 47.2. Considera o conjunto dos pontos $P(x, y)$ que satisfazem a condição $\overrightarrow{A B} \cdot \overrightarrow{B P}=0$ . Representa essa condição por uma equação e resolve-a em ordem a $y$ . 140

Studdy Solution

STEP 1

1. The circle is centered at $A(0,3)$ and passes through $B(3,-1)$ .
2. The line $r$ is tangent to the circle at point $B$ .
3. The vector condition $\overrightarrow{A B} \cdot \overrightarrow{B P}=0$ defines a set of points $P(x, y)$ .

STEP 2

1. Determine the radius of the circle.
2. Find the equation of the tangent line $r$ at point $B$ .
3. Represent the vector condition as an equation.
4. Solve the equation for $y$ .

STEP 3

Determine the radius of the circle using the distance formula between points $A(0,3)$ and $B(3,-1)$ :
$\text{Radius} = \sqrt{(3-0)^2 + (-1-3)^2}$ $= \sqrt{3^2 + (-4)^2}$ $= \sqrt{9 + 16}$ $= \sqrt{25}$ $= 5$

STEP 4

Find the slope of the line perpendicular to $\overline{AB}$ (since the tangent is perpendicular to the radius at the point of tangency):
1. Calculate the slope of $\overline{AB}$ :
$\text{Slope of } \overline{AB} = \frac{-1 - 3}{3 - 0} = \frac{-4}{3}$
2. The slope of the tangent line $r$ is the negative reciprocal:
$\text{Slope of } r = \frac{3}{4}$
3. Use the point-slope form to find the equation of the tangent line $r$ at point $B(3, -1)$ :
$y + 1 = \frac{3}{4}(x - 3)$
4. Simplify to the slope-intercept form:
$y = \frac{3}{4}x - \frac{9}{4} - 1$ $y = \frac{3}{4}x - \frac{13}{4}$

STEP 5

Represent the vector condition $\overrightarrow{A B} \cdot \overrightarrow{B P}=0$ :
1. Calculate vector $\overrightarrow{A B} = (3 - 0, -1 - 3) = (3, -4)$ .
2. Let $P(x, y)$ be a point, then $\overrightarrow{B P} = (x - 3, y + 1)$ .
3. The dot product condition is:
$\overrightarrow{A B} \cdot \overrightarrow{B P} = 3(x - 3) + (-4)(y + 1) = 0$

STEP 6

Solve the equation for $y$ :
1. Expand and simplify:
$3(x - 3) - 4(y + 1) = 0$ $3x - 9 - 4y - 4 = 0$ $3x - 4y - 13 = 0$
2. Solve for $y$ :
$-4y = -3x + 13$ $y = \frac{3}{4}x - \frac{13}{4}$
The equation of the tangent line $r$ is: $y = \frac{3}{4}x - \frac{13}{4}$
The equation representing the vector condition is: $y = \frac{3}{4}x - \frac{13}{4}$

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