Question

A circle of radius square root of 10 touches the line x - 3y = 2. Find the general equation of the locus of its center. (Two solutions)

Find the general equations of the bisectors of the angles between the lines:
a. 3x + y = 6 and x + 3y = -2
b. 4x - 5y = 26 and 5x + 4y = 20

Find the general equation of the line through the point A(-1, 2) and passing at a distance 3 from the point Q(-10, -5).

A circle of radius 6 touches both the coordinate axes. A line with slope -3/4 passes over and just touches the circle. If the circle is in the first quadrant, find the general equation of the line.

Answer 1

Answer on Question #57673, Math/ Analytic Geometry

A circle of radius square root of 10 touches the line $x - 3y = 2$. Find the general equation of the locus of its center. (Two solutions)

Solution: $r = \sqrt{10}$; O $(x, y)$-center of the such circle.

The locus of the center such circles is two lines which is parallel to the line $x - 3y = 2$. The distances between new lines and given line is square root of 10.

The line $x - 3y = 2$ through the point $A(2,0)$ (for example),

The general equation for the parallel line is $x - 3y = C$; C-some constant, so O belongs to $x - 3y = C$.

The distance from A to the line $x - 3y = C$ is $r = \sqrt{10}$, so find C from this condition.

$$\text{distance}(ax + by + c = 0, (x_0, y_0)) = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}.$$

We get

$$\sqrt{10} = \frac{|2 - C|}{\sqrt{1^2 + (-3)^3}} = \frac{|2 - C|}{\sqrt{10}}$$

$|2 - C| = 10$; $C_1 = 12$; $C_2 = -8$;

The equation of the locus $L_1$: $x - 3y = 12$; $L_2$: $x - 3y = -8$;

Answer: $L_1$: $x - 3y = 12$; $L_2$: $x - 3y = -8$.

Find the general equations of the bisectors of the angles between the lines:

a. $3x + y = 6$ and $x + 3y = -2$

b. $4x - 5y = 26$ and $5x + 4y = 20$

Solution:

The equations of the bisectors of the angles between the lines $a_1x + b_1y + c_1 = 0$;

$$a_2x + b_2y + c_2 = 0:$$ $$\frac{a_1x + b_1y + c_1}{\sqrt{a_1^2 + b_1^2}} = \pm \frac{a_2x + b_2y + c_2}{\sqrt{a_2^2 + b^2}}$$

a. $3x + y = 6$ and $x + 3y = -2$;

$$3x + y - 6 = 0 \text{ and } x + 3y + 2 = 0;$$ $$\frac{3x + y - 6}{\sqrt{3^2 + 1^2}} = \pm \frac{x + 3y + 2}{\sqrt{1^2 + 3^2}}; \quad \frac{3x + y - 6}{\sqrt{10}} = \pm \frac{x + 3y + 2}{\sqrt{10}};$$ $$3x + y - 6 = x + 3y + 2;$$ $$2x - 2y - 8 = 0$$ $$x - y - 4 = 0;$$ $$3x + y - 6 = -(x + 3y + 2);$$ $$3x + y - 6 = -x - 3y - 2;$$ $$4x + 4y - 4 = 0;$$ $$x + y - 1 = 0;$$

b. $4x - 5y = 26$ and $5x + 4y = 20$;

$$4x - 5y - 26 = 0 \text{ and } 5x + 4y - 20 = 0;$$ $$\frac{4x - 5y - 26}{\sqrt{4^2 + (-5)^2}} = \pm \frac{5x + 4y - 20}{\sqrt{5^2 + 4^2}}; \quad \frac{4x - 5y - 26}{\sqrt{41}} = \pm \frac{5x + 4y - 20}{\sqrt{41}}$$

4x - 5y -26 = 5x + 4y - 20;

-x-9y-6=0;

x+9y+6=0;

4x - 5y -26 = -(5x + 4y - 20);

4x - 5y -26 = -5x - 4y + 20;

9x - y - 46 = 0;

Answer: a) x - y - 4 = 0; x + y - 1 = 0; b) x + 9y + 6 = 0; 9x - y - 46 = 0.

Find the general equation of the line through the point A(-1, 2) and passing at a distance 3 from the point Q(-10, -5).

Solution: The general equation of the line through the point A(-1, 2): y - 2 = k(x + 1);

-kx + y - k - 2 = 0;

distance(ax + by + c = 0, (x₀, y₀)) = |ax₀ + by₀ + c| / √a² + b².

3 = distance(y - kx - k - 2; Q(-10, -5)) = |10k - 5 - k - 2| / √(-k)² + 1² = |9k - 7| / √k² + 1²;

81k^2 + 1 = 3; squaring the expression: (9k - 7)² = 9(k² + 1);

81k^2 - 126k + 49 = 9k^2 + 9;

72k^2 - 126k + 40 = 0;

36k^2 - 63k + 20 = 0;

D = (-63)^2 - 4*36*20 = 3969 - 2880 = 1089 = 33^2;

k₁ = 63 + 33 / 72 = 4/3;

k₂ = 63 - 33 / 72 = 5/12;

L₁: -4/3 x + y - 4/3 -2 = 0; simplifying the equation we get L₁: -4x + 3y - 10 = 0;

L₂: -5/12 x + y - 5/12 -2 = 0; simplifying the equation we get L₂: -5x + 12y - 29 = 0;

Answer: -4x + 3y - 10 = 0; -5x + 12y - 29 = 0;

A circle of radius 6 touches both the coordinate axes. A line with slope -3/4 passes over and just touches the circle. If the circle is in the first quadrant, find the general equation of the line.

Solution. The equation of the line with slope -3/4 is y = -3/4 x + b; If the circle is in the first quadrant we get that its center is O(6,6); The distance from O to the line y = -3/4 x + c equals the radius 6, rewrite y = -3/4 x + c as 4y + 3x - 4c = 0;

$$\text{distance}(ax + by + c = 0, (x_0, y_0)) = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}.$$ $$6 = \frac{|a x_0 + b y_0 + c|}{\sqrt{a^2 + b^2}} = \frac{|3*6 + 4*6 - 4c|}{\sqrt{3^2 + 4^2}} = \frac{|18 + 24 - 4c|}{\sqrt{25}}$$ $$6 = \frac{|42 - 4c|}{5}; \quad |42 - 4c| = 30; \quad c = 3; \quad c = 18; \quad L_1: 3x + 4y - 12 = 0; \quad L_2: 3x + 4y - 72 = 0,$$

but $L_2: 3x + 4y - 72 = 0$ passes over circle.

Answer: $3x + 4y - 72 = 0$.

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