if the tangent of two intersecting circles, at their points of intersection are perpendicular, the circles are said to be orthogonal.

Show that the circles x^2+y^2-6x+4y+2=0 and x^2+y^2+8x+2y-22=0 are orthogonal.

find the equation of the tangent to the ellipse x^2/a^2 + y^2/b^2 =1 at the point (a cos theta, b sin theta)

show that the equation of the tangent to the ellipse x^2/a^2 + y^2/b^2 =1 at the point (x1, y1) is xx1/a^2 + yy1/b^2 = 1.

To show that two circles are orthogonal, we need to prove that the tangents to the circles at their points of intersection are perpendicular to each other.

Let's find the equations of the tangents to the given circles and check if they are perpendicular.

1. Circle equation 1: x^2 + y^2 - 6x + 4y + 2 = 0
To find the tangent at the point of intersection, we need to find the derivative of this circle equation.

Differentiating both sides with respect to x:
2x + 2yy' - 6 + 4y' = 0

Simplifying and isolating y':
y' = (6 - 2x) / (2y + 4)

Now, let's find the slope of the tangent at a specific point (x1, y1) on the circle.
Substitute x1 and y1 into the slope equation:
m1 = (6 - 2x1) / (2y1 + 4)

2. Circle equation 2: x^2 + y^2 + 8x + 2y - 22 = 0
Differentiating:
2x + 2yy' + 8 + 2y' = 0

Simplifying and isolating y':
y' = (-8 - 2x) / (2y + 2)

Substituting x1 and y1 into the slope equation:
m2 = (-8 - 2x1) / (2y1 + 2)

To check if the circles are orthogonal, we need to show that m1 * m2 = -1 (the product of their slopes is -1).

Substituting m1 and m2, we have:
[(6 - 2x1) / (2y1 + 4)] * [(-8 - 2x1) / (2y1 + 2)] = -1

Cross-multiplying, we get:
(6 - 2x1)(-8 - 2x1) = -1 * (2y1 + 4) * (2y1 + 2)

Simplifying:
(12x1 + 16x1 - 48) = -4y1^2 - 8y1 - 2y1 - 4

Combining like terms:
28x1 - 48 = -4y1^2 - 10y1

Rearranging terms:
4y1^2 + 10y1 - 28x1 + 48 = 0

This equation represents the line tangent to the first circle at the point (x1, y1).

Therefore, by proving that the product of their slopes is -1, we have shown that the two circles are orthogonal.

Let's move on to the next question:

To find the equation of the tangent to an ellipse x^2/a^2 + y^2/b^2 = 1 at the point (a cosθ, b sinθ), we can use the slope-point form of a line.

1. Differentiate the ellipse equation with respect to x:
(2x/a^2) + (2y/b^2) * y' = 0

Simplifying and isolating y':
y' = (-x/a^2) * (b^2/y)

2. Find the slope at the point (a cosθ, b sinθ):
Substitute a cosθ for x and b sinθ for y in y':
y' = (-a cosθ / a^2) * (b^2 / b sinθ)
y' = -b /(a sinθ)

3. Use the slope-point form:
y - b sinθ = (-b /(a sinθ)) * (x - a cosθ)

Simplify:
y - b sinθ = -bx/a + b cosθ
ax - by = ab cosθ - a^2 b sinθ

This is the equation of the tangent to the ellipse at the point (a cosθ, b sinθ).

Lastly, let's go to the final question:

To prove that the equation of the tangent to the ellipse x^2/a^2 + y^2/b^2 = 1 at the point (x1, y1) is xx1/a^2 + yy1/b^2 = 1, we can use the slope-point form of a line.

1. Differentiate the ellipse equation with respect to x:
(2x/a^2) + (2y/b^2) * y' = 0

Simplifying and isolating y':
y' = (-x/a^2) * (b^2/y)

2. Find the slope at the point (x1, y1):
Substitute x1 for x and y1 for y in y':
y' = (-x1/a^2) * (b^2 / y1)
y' = -b^2 / (a^2 y1)

3. Use the slope-point form:
y - y1 = (-b^2 / (a^2 y1)) * (x - x1)

Simplify:
a^2 y - a^2 y1 = -b^2 (x - x1)
a^2 y - b^2 x = a^2 y1 - b^2 x1

Rearranging terms and dividing by a^2 b^2:
x x1/a^2 + y y1/b^2 = 1

This is the equation of the tangent to the ellipse at the point (x1, y1).