Question

Find the equation of the circle determined by the given conditions:
10. tangent to the x-axis and passing through (5, 1) and (12, 8)
11. tangent to the line y= -1 and containing the points (1, 1) and (3, 3)
12. tangent to the lines 2x + y = 4, 2x + y =2 and x - 2y + 5 = 0
13. passing through the origin and tangent to the lines 3x + 4y - 10 = 0 and 4x + 3y - 5 = 0
14. inscribed in a triangle with sides on the lines x - 3y = -5, 3x + y = 1 and 3x - y = -11
15. inscribed in a triangle whose vertices are (0, 6), (8, 6), and (0, 0)
16. having radius of square root of 5, through (0, 4) and (3, 7).
17. radius and tangent to the line 2x + y -1 = 0 at (1, -1)
18. tangent to 3x - 2y - 9 = 0 and 2x - 3y - 1 = 0 and center on 2x + y - 10 = 0

Answer 1

Answer on Question #57676 – Math – Analytic Geometry

Question

Find the equation of the circle determined by the given condition:

10. tangent to the x-axis and passing through (5,1) and (12,8).

Solution

Equation of circle $(x - a)^2 + (y - b)^2 = R^2$, where $(a, b)$ is the center of the circle, $R$ is the radius of the circle. Hence

$$\left\{ \begin{array}{c} (5 - a)^2 + (1 - b)^2 = R^2 \\ (12 - a)^2 + (8 - b)^2 = R^2 \\ |b| = R \end{array} \right.$$

Solutions of the system are $a_1 = 0$, $b_1 = 13$ and $a_2 = 8$, $b_2 = 5$.

Equations of circle are $x^2 + (y - 13)^2 = 13^2$ and $(x - 8)^2 + (y - 5)^2 = 5^2$.

**Answer**: $x^2 + (y - 13)^2 = 13^2$, $(x - 8)^2 + (y - 5)^2 = 5^2$.

Question

Find the equation of the circle determined by the given condition:

11. tangent to the line $y = -1$ and containing the point (1,1) and (3,3).

Solution

Equation of the circle is $(x - a)^2 + (y - b)^2 = R^2$, where $(a, b)$ is the center of the circle, $R$ is the radius of the circle. Hence

$$\left\{ \begin{array}{c} (1 - a)^2 + (1 - b)^2 = R^2 \\ (3 - a)^2 + (3 - b)^2 = R^2 \\ |b + 1| = R \end{array} \right.$$

Solutions of the system are $a_1 = 3$, $b_1 = 1$ and $a_2 = -5$, $b_2 = 9$.

Equations of circle are $(x - 3)^2 + (y - 1)^2 = 2^2$ and $(x + 5)^2 + (y - 9)^2 = 10^2$.

**Answer**: $(x - 3)^2 + (y - 1)^2 = 2^2$, $(x + 5)^2 + (y - 9)^2 = 10^2$.

Question

Find the equation of the circle determined by the given condition:

12. tangent to the line $2x + y = 4$, $2x + y = 2$ and $x - 2y + 5 = 0$.

Solution

Equation of circle is $(x - a)^2 + (y - b)^2 = R^2$, where $(a, b)$ is the center of the circle, $R$ is the radius of the circle. Hence

$$\left\{ \begin{array}{l} \frac{|2a + b - 4|}{\sqrt{5}} = R \\ \frac{|2a + b - 2|}{\sqrt{5}} = R \\ \frac{|a - 2b + 5|}{\sqrt{5}} = R \end{array} \right.$$

Solutions of the system are $a_1 = \frac{2}{5}$, $b_1 = \frac{11}{5}$ and $a_2 = 0$, $b_2 = 3$.

Equations of the circle are $(x - \frac{2}{5})^2 + (y - \frac{11}{5})^2 = \frac{1}{5}$ and $x^2 + (y - 3)^2 = \frac{1}{5}$.

**Answer**: $(x - \frac{2}{5})^2 + (y - \frac{11}{5})^2 = \frac{1}{5}$, $x^2 + (y - 3)^2 = \frac{1}{5}$.

Question

Find the equation of the circle determined by the given condition:

13. passing through the origin and tangent to the line $3x + 4y - 10 = 0, 4x + 3y - 5 = 0$

Solution

Equation of circle $(x - a)^2 + (y - b)^2 = R^2$, where $(a, b)$ is the center of the circle, $R$ is the radius of the circle. Hence

$$\left\{ \begin{array}{l} a ^ {2} + b ^ {2} = R ^ {2} \\ \frac {| 3 a + 4 b - 1 0 |}{5} = R \\ \frac {| 4 a + 3 b - 5 |}{5} = R \end{array} \right.$$

The system of equations has no solution. Thus, such a circle does not exist.

Answer: Circle does not exist.

Question

Find the equation of the circle determined by the given condition:

14. inscribed in a triangle with sides on the lines $x - 3y = -5$, $3x + y = 1$, $3x - y = -11$.

Solution

Equation of circle $(x - a)^2 + (y - b)^2 = R^2$, where $(a, b)$ is the center of the circle, $R$ is the radius of the circle. Hence

$$\left\{ \begin{array}{l} \frac {| a - 3 b + 5 |}{\sqrt {1 0}} = R \\ \frac {| 3 a + b - 1 |}{\sqrt {1 0}} = R \\ \frac {| 3 a - b + 1 1 |}{\sqrt {1 0}} = R \end{array} \right.$$

Solution of the system is $a = \frac{-5}{3}$, $b = \frac{7}{3}$. Equation of the circle is $(x + \frac{5}{3})^2 + \left(y - \frac{7}{3}\right)^2 = \frac{121}{90}$.

Answer: $(x + \frac{5}{3})^2 + \left(y - \frac{7}{3}\right)^2 = \frac{121}{90}$

Question

Find the equation of the circle determined by the given condition:

15. inscribed in a triangle with vertices (0,6), (8,6), (0,0).

Solution

Equation of the circle is $(x - a)^2 + (y - b)^2 = R^2$, where $(a, b)$ is the center of the circle, $R$ is the radius of the circle. The equations of lines passing through the vertices of the triangle are $x = 0$, $y = 6$, $6x - 8y = 0$. Hence

$$\left\{ \begin{array}{l} \frac {| a |}{1} = R \\ \frac {| 6 - b |}{1} = R \\ \frac {| 6 a - 8 b |}{1 0} = R \end{array} \right.$$

Solution of the system is $a = 2$, $b = 4$. Equation of the circle is $(x - 2)^2 + (y - 4)^2 = 4$.

Answer: $(x - 2)^{2} + (y - 4)^{2} = 4$

Question

Find the equation of the circle determined by the given condition:

16. having radius of square root of 5, through (0,4) and (3,7).

Solution

Equation of the circle is $(x - a)^2 + (y - b)^2 = R^2$, where $(a, b)$ is the center of the circle, $R$ is the radius of the circle. Hence

$$\left\{ \begin{array}{c} a ^ {2} + (4 - b) ^ {2} = 5 \\ (3 - a) ^ {2} + (7 - b) ^ {2} = 5 \end{array} \right.$$

Solutions of the system are $a_1 = 1$, $b_1 = 6$ and $a_2 = 2$, $b_2 = 5$.

Equations of the circle are $(x - 1)^2 + (y - 6)^2 = 5$ and $(x - 2)^2 + (y - 5)^2 = 5$.

Answer: $(x - 1)^2 + (y - 6)^2 = 5$, $(x - 2)^2 + (y - 5)^2 = 5$.

Question

Find the equation of the circle determined by the given condition:

17. radius and tangent to the line $2x + y - 1 = 0$ at (1,-1).

Solution

Equation of the circle is $(x - a)^2 + (y - b)^2 = R^2$, where $(a, b)$ is the center of the circle, $R$ is the radius of the circle.

Line perpendicular to $2x + y - 1 = 0$ at (1, -1) is $\frac{x - 1}{2} = \frac{y + 1}{1}$, hence $x - 2y - 3 = 0$ and $a - 2b - 3 = 0$, that is, $a = 2b + 3$.

Given $R = \sqrt{(a - 1)^2 + (b + 1)^2}$, hence

$$R = \sqrt {(2 b + 3 - 1) ^ {2} + (b + 1) ^ {2}} = \sqrt {(2 b + 2) ^ {2} + (b + 1) ^ {2}} = \sqrt {5 b ^ {2} + 1 0 b + 5} = \sqrt {5} | b + 1 |,$$

that is, $|b + 1| = \frac{R}{\sqrt{5}}$.

Thus, $b_{1} = -1 - \frac{R}{\sqrt{5}}$ or $b_{2} = -1 + \frac{R}{\sqrt{5}}$, hence $a_{1} = 2b_{1} + 3 = 2\left(-1 - \frac{R}{\sqrt{5}}\right) + 3 = 1 - \frac{2R}{\sqrt{5}}$ or $a_{2} = 2b_{2} + 3 = 2\left(-1 + \frac{R}{\sqrt{5}}\right) + 3 = 1 + \frac{2R}{\sqrt{5}}$.

Equations of the circle are $(x - 1 + \frac{2R}{\sqrt{5}})^2 + \left(y + 1 + \frac{R}{\sqrt{5}}\right)^2 = R^2$ and

$$\left(x - 1 - \frac {2 R}{\sqrt {5}}\right) ^ {2} + \left(y + 1 - \frac {R}{\sqrt {5}}\right) ^ {2} = R ^ {2}.$$

Answer: $(x - 1 + \frac{2R}{\sqrt{5}})^2 + \left(y + 1 + \frac{R}{\sqrt{5}}\right)^2 = R^2$, $(x - 1 - \frac{2R}{\sqrt{5}})^2 + \left(y + 1 - \frac{R}{\sqrt{5}}\right)^2 = R^2$.

Question

Find the equation of the circle determined by the given condition:

18. tangent to $3x - 2y - 9 = 0$ and $2x - 3y - 1 = 0$ and center on $2x + y - 10 = 0$.

Solution

Equation of circle is $(x - a)^2 + (y - b)^2 = R^2$, where $(a, b)$ is the center of the circle, $R$ is the radius of the circle.

$$\left\{ \begin{array}{l} 2 a + b - 1 0 = 0 \\ \frac {| 3 a - 2 b - 9 |}{\sqrt {1 3}} = R \\ \frac {| 2 a - 3 b - 1 |}{\sqrt {1 3}} = R \end{array} \right.$$

Solutions of the system are $a_1 = 2$ , $b_1 = 6$ and $a_2 = 4$ , $b_2 = 2$ .

Equations of the circle are $(x - 2)^{2} + (y - 6)^{2} = \frac{225}{13}$ and $(x - 4)^{2} + (y - 2)^{2} = 1$ .

Answer: $(x - 2)^{2} + (y - 6)^{2} = \frac{225}{13}$ , $(x - 4)^{2} + (y - 2)^{2} = 1$ .

www.AssignmentExpert.com