Have you left out something? There is a different tangent to the circle at every point on the circle. Which one are you seeking? Given a point on the circle, the tangent line there is perpendicular to the radius.2)a) find the equation of the tangent to the circle x^2+y^2-4x+6y-7=0
You can find the center of the given circle by completing the square: , . That tells you that the center of the circle is at (2, -3). What is the slope of the line through (2, -3) and ? What will the slope of the perpendicular line be? What is the equation of the line perpendicular to the radius through ?
Once you know the equation of the tangent line, set x= 0 and solve for y, then set y= 0 and solve for x.b)given that this tangent meets the x axis at point a and y axis at B find the coordaniates of A and B
You don't need calculus for this problem, just geometry and algebra.3)show that y=2x+3 is a tangent to the circle x^2+y^2-12x-15y+81=0
(dont even have a clue here, i can find the equation of the tangent given the point where is crosses, but from the equation you cant work out gradient or point where it crosses)
In order to be a "tangent" the line must cut the circle in exactly one point- that is there exist one (x, y) which satisfies both equations. Since y= 2x+ 3, you can substitute 2x+ 3 for y in the equation of the circle. . . That equation will have a single solution (and the line will be tangent to the circle) if and only if the left side is a perfect square. Is it?
Hopefully, you found a single value of x satisfying the equation above. Since it lies on the line y= 2x+ 3, (x, 2x+ 3) is that point of tangency.b)find the points where the tangent touches the circle
Thankyou in advanced