Find the equation of the circle tangent to the circle x²+y²+2x-6y+5=0 at (1,2), and passing through (4,1)?

Question

so basically you have an unknown circle that passes through (1,2) and (4,1), and is tangent to another circle x²+y²+2x-6y+5=0 at that point (1,2).

[x²+y²+2x-6y+5=0 has a center of (-1, 3) and a radius of √5.]

i know how to solve for a circle with 3 points; am I supposed to look for a third point? or is the fact that the circle is tangent to another circle even more relevant than it seems?

that's all that's given, it cannot be impossible, and i've seen something like it on Answers. immediate help would be gladly gladly appreciated.

@ ScarletManuka: i have no doubts your answer is right, i'm just not seeing how

"Since the first circle has centre (-1, 3) and the tangent point is (1, 2), we know the second circle has a centre of form
(-1+t, 3+2t) for some t. "

i think i may have forgotten some valuable geometry/algebra principle. can you show why the above is so?

We need this to be equidistant from (1, 2) and (4, 1), so we get
(-2+t)^2 + (1+2t)^2 = (-5+t)^2 + (2+2t)^2

why this, too?

even if this is Answers, sorry for the inconvenience. ><

Answer 1

If two circles are tangent to each other, then the line joining their centres passes through the tangent point. (Another way to think about this is that if we draw the line from each centre to the tangent point, the two halves will form a straight line; that is because we know each half is at right angles to the tangent line.)

Since the first circle has centre (-1, 3) and the tangent point is (1, 2), we know the second circle has a centre of form
(-1+t, 3+2t) for some t. We need this to be equidistant from (1, 2) and (4, 1), so we get

(-2+t)^2 + (1+2t)^2 = (-5+t)^2 + (2+2t)^2
<=> 4 - 4t + t^2 + 1 + 4t + 4t^2 = 25 - 10t + t^2 + 4 + 8t + 4t^2
<=> 5 + 5t^2 = 29 - 2t + 5t^2
<=> 2t = 24
So t = 12 and the centre of the circle is (11, 27). Its radius is √(10^2 + 25^2) = √725 = 5√29. So its equation is
(x-11)^2 + (y-27)^2 = 725
<=> x^2 + y^2 - 22x - 54y + 125 = 0.

Answer 2

Set x = 0. in case you get in basic terms one attainable answer for x, then the y-axis touches as quickly as and is a tangent. x^2 + y^2 - 2x - 10y + 25 = 0 0^2 + y^2 - 2(0) - 10y + 25 = 0 y^2 - 10y + 25 = 0 are you able to think of of two numbers that multiply to grant 25 and upload to -10? i admire -5 and -5. specific, it is the comparable variety two times, yet that keeps to be 2 numbers. (y - 5)(y - 5) = 0 subsequently, y-5 = 0 y = 5 The circle touches the y-axis in basic terms as quickly as at (0, 5). subsequently, the y-axis is a tangent to the circle.