60) find the equations of the two tangents to the circle x^2 + 4x +y^2 - 6y - 3 = 0 which pass through (-2,1)?

Answer 1

The equation of the circle is:

x² + 4x + y² - 6y - 3 = 0
x² + 4x + y² - 6y = 3

Complete the square for x and y.

x² + 4x + 4 + y² - 6y + 9 = 3 + 4 + 9
(x² + 4x + 4) + (y² - 6y + 9) = 16
(x + 2)² + (y - 3)² = 16

This is a circle with center (-2, 3) and radius 4. The distance between the center and the point (-2, 1) is 2. Since this is less than the radius the point (-2, 1) is inside the circle. No tangent line can be drawn from a point inside the circle to the circle.

There is no solution.

Answer 2

Hi Allison,

If you rewrite the circle's equation into standard form, you get::

x^2 + 4x +y^2 - 6y - 3 = 0
x^2 + 4x + 4 + y^2 - 6y + 9 = 3 + 4 + 9
(x + 2)^2 + (y - 3)^ 2 = 16

This tells you the circle's center is at (-2,3). Since your other given point is at (-2,1), it is 2 units away from the center of the circle. This means that the point (-2,1) is actually INSIDE the circle, so you can't draw tangents to the circle from that point!

I hope this helps! Otherwise check your point's coordinates. This point just doesn't make any sense.

:-) Have a nice day!!

Answer 3

A normal way to solve this is as follows:
Complete the square:
(x+2)² + (y-3)² = 4²
So implicit differentiation gives
2(x+2) + 2(y-3)(dy/dx) = 0

dy/dx = (x+2)/(y-3)

The equation of the line through (-2,1) will be
(y-1) = m(x+2) where that m has to match the dy/dx.

However, the circle above has center at -2, 3 with radius 4.
(-2,1) is in the interior of this circle, so it is mighty hard for a tangent to pass through it.

Answer 4

x^2+y^2-4x+3=0 could be writen: (x - 2)^2 + y^2 - one million = 0 (one million) equation of a circle with center (2, 0) and radius = one million, intersecting OX axis at x = one million and x = 3 x^2+y^2 + ok = 0 equation of a circle with center in beginning and tangent to (one million) if: ok = -one million (tangent exterior) ok = -9 (tangent inner)

Answer 5

you hurt my brain i hope your happy.