Tangent of a circle lesson

Question

I'm confuse on solving this problem:

Find the equation of the circle which is tangent to the line $3x-2y=5$ at $(3,2)$ and it is passing through $(-2,1)$.

Can you pretty answer and explain how did you do the solution?

Please. I barely need to understand this lesson a lot better. Thank you.

When you write "Touching", do you mean "tangentially" ? – DonAntonio Dec 07 '16 at 07:54 — DonAntonio, Dec 07 '16 at 07:54

Robert Z · Answer 1 · 2016-12-07T13:22:42.540

1

Hint. Let $(x_0,y_0)$ be the centre of such circle. Then

i) $(x_0,y_0)$ stays on the line orthogonal to $3x-2y=5$ at $(3,2)$: $$2(x-3)+3(y-2)=0$$ that is $$2x+3y=12.$$ ii) $(x_0,y_0)$ stays on the perpendicular bisector of the segment of extreme points $(3,2)$ and $(-2,1)$: $$(x-3)^2+(y-2)^2=(x+2)^2+(y-1)^2$$
that is $$5x+y=4$$ Can you take it from here?

edited Dec 07 '16 at 13:22

answered Dec 07 '16 at 07:55

Robert Z

145,942

Can you help me? How to solve it? – Trixie Dec 07 '16 at 07:57
@Trixie Are you able to find the equation of the line in i)? – Robert Z Dec 07 '16 at 08:06
What common property do all points have that lie on a bisector of a line segment? – H1ghfiv3 Dec 07 '16 at 08:10

Tangent of a circle lesson

1 Answers1