Question

1. Find the equation of the image of the circle x² + y2 + 16x-24y + 183 = 0 by rotated the line mirror 4x + 7y + 13 = 0. 2. The image of the circle (x - 3)² + (y-2)² = 1 in the line mirror ax + by = 19 is (x-1)³ + (y-16)2 = 1 then, find the values of (a, b). 3. Find the equation of a line passing through the origin and making an angle with the 4 line y-3x-5. 4. A parabola is drawn with its focus at (3,4) and vertex at the focus of the parabola y²-12x - 4y + 4 = 0. The n find equation of the parabola. 5. If the line ax + by + c = 0 touches the circle x² + y² - 2x = and is normal to the circle x² + y² + 2x - 4y + 1 = 0, then find the value of (a, b). 6. If the line through the points (-2, 6) and (4, 8) is perpendicular to the line through the points (8, 12) and (x, 24). Find the value of x. -3 7.1² 14 231= [] then find the matrix A 8. Find the equation of the ellipse having its center at the point (2,-3), one and one vertex at (4, -3). 3 9. Find the value of x if-1 0 10. Solve the linear system using Cramer's rule a) 2 1 2 4 (6x - 4y = -12 8x - 3y = -2 X = 16 -21 3x + 2y = z = 5 b) x-y+3z = -15 (2x + y +7z = -28 one focus at (3,-3) 11. Find the value of k for which the following system of linear equations has infinite solutions: x + (k+1)y = 5 ((k+1)x + 9y = 8k - 1​

Answer 1

-72x - 53y + 287 = 0.

Explanation:

To find the equation of the image of the circle, we need to reflect each point on the circle in the given line mirror.

The line mirror equation is given as 4x + 7y + 13 = 0.

The reflection of a point (x, y) in the line mirror can be found using the formula:

x' = (x - 2Ay - 2B(Ax + By + C)) / (A^2 + B^2)

y' = (y - 2Bx + 2A(Ax + By + C)) / (A^2 + B^2)

where A, B, and C are the coefficients of the line mirror equation.

For the given line mirror equation 4x + 7y + 13 = 0, we have A = 4, B = 7, and C = 13.

Now, let's find the equations of the image of the circle.

The original circle equation is x² + y² + 16x - 24y + 183 = 0.

Using the reflection formulas, we substitute the values of x and y in the circle equation to find x' and y':

x' = (x - 2Ay - 2B(Ax + By + C)) / (A^2 + B^2)

= (x - 2(4)y - 2(7)(4x + 7y + 13)) / (4^2 + 7^2)

= (x - 8y - 8(4x + 7y + 13)) / 65

= (x - 8y - 32x - 56y - 104) / 65

= (-31x - 64y - 104) / 65

y' = (y - 2Bx + 2A(Ax + By + C)) / (A^2 + B^2)

= (y - 2(7)x + 2(4)(Ax + By + C)) / (4^2 + 7^2)

= (y - 14x + 8(Ax + By + C)) / 65

= (y - 14x + 8(4x + 7y + 13)) / 65

= (57x + 35y + 104) / 65

Therefore, the equation of the image of the circle is:

(-31x - 64y - 104) / 65 + (-57x + 35y + 104) / 65 + 16x - 24y + 183 = 0

Simplifying the equation, we get:

-31x - 64y - 57x + 35y + 16x - 24y + 183 + 104 = 0

-72x - 53y + 287 = 0

So, the equation of the image of the circle is -72x - 53y + 287 = 0.