1.)for the circle x^2+y^2+6x-4y+3=0
find the equation of the tangent line at (-2,5)
2.)a circle is tangent to the y-axis at y=3 and has one x-intercept at x=1
determine the other x-int and deduce the equation of the circle
any help or hints on how to solve this problem would be GREATLY appreciated
Thanks in advance
thanks so much alexmahone but im still stumped at the second problem
btw i did it a bit diff way
i found the slope of (-2,5) to (-3,2)<-- Center
and found out it was m=3
so reciprical of that is -1/3
and then i plugged it into the point slope form and got
y-5=-1/3(x+2)
so y=-x/3+13/3
right?
here is the solution
let the coordinate of other intercept be (a,0) (its on x axis)
the coordinate of middle point of (1,0) and (a,0) will be ((a+1)/2,0)(let us name it as A).its magnitude will be equal to the length of radius.
a line drawn perpendicular to this middle point and y intercept will intersect at centre of the circle (let centre of circle be O).
now it can be easily observed that coordinate of centre of circle will be ((a+1)/2,3)
let intercept (1,0) be B
so triangle AOB is a right angle triangle where AO=3 (length of y intercept),OB=length of radius=(a+1)/2,AB=[(a+1)/2]-1=(a-1)/2
now applying pythagores theorem we have
[(a+1)/2]^2=[(a-1)/2]^2+9
solving this we get a=9
so the coordinate of other intercept will be (9,0) and radius will be 5 units snd coordinate of centre will be (5,3)
and equation of circle will be
(x-5)^2+(y-3)^2=25 or
x^2+y^2-10x-6y+9=0
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Originally Posted by saar4ever 
