Question

show that x² + y² - 6x - 2y + 1 = 0, x² + y² + 2x - 8y + 13 =0 circles touch each other. Find the point of contact and the equation of common tangent at their point of contact.​

Answer 1

The equation of the common tangent at the point of contact (1, -1/2) is $y = 2x - 1/2$

The given equations represent two circles:

x² + y² - 6x - 2y + 1 = 0

x² + y² + 2x - 8y + 13 = 0

To find the point of contact of these two circles, we need to find their point of intersection. Setting the x and y values equal in both equations, we get:

x² + y² - 6x - 2y + 1 = x² + y² + 2x - 8y + 13

Solving for x and y, we find:

-8x + 6y = 12

$x = 3/4 + 3/4y$

Substituting the value of x in the first equation, we get:

$(3/4 + 3/4y)$² + y² - $6(3/4 + 3/4y) - 2y + 1 = 0$

Expanding and solving for y, we get:

$y = -1/2$

Substituting this value of y back into $x = 3/4 + 3/4y$, we get:

So the point of contact of the two circles is (1, -1/2).

To find the equation of the common tangent at this point of contact, we differentiate both the equations and equate the slopes of the tangents at (1, -1/2). T

he slope of the tangent to a circle at (x, y) is given by:

$-2x/2y$

So the slope of the tangent to the first circle at (1, -1/2) is:

$-2(1)/(-2(-1/2)) = 2$

The slope of the tangent to the second circle at (1, -1/2) is:

$-2(1)/(-2(-1/2)) = 2$

Since the slopes are equal, the tangents are parallel and therefore have the same equation.

The equation of the line passing through (1, -1/2) with slope 2 is:

$y - (-1/2) = 2(x - 1)$

$y = 2x - 1/2$

So the equation of the common tangent at the point of contact (1, -1/2) is:

$y = 2x - 1/2$

More questions and answers

brainly.in/question/12321085

https://brainly.in/question/54137822

#SPJ1