# 2.4.10 3 circles one intersection

• MHB
• karush
In summary, the equations (x+4)^2+(y+11)^2=169, (x-9)^2+(y+5)^2=100, and (x-4)^2+(y-5)^2=25 represent three intersecting circles. By setting the equations equal to each other, it is possible to find the equation of the line passing through the intersection points. This line can be used to find the intersection point of the circles, which is (1,1). This method can also be used to find the equation of a chord joining the intersecting points. The point of intersection of three circles can be found using basic algebra, rather than a limit procedure.

#### karush

Gold Member
MHB
$\tiny{\textbf{2.4.10}}$
$\begin{array}{rl} (x+4)^2+(y+11)^2&=169 \\ (x-9)^2+(y+5)^2&=100 \\ (x-4)^2+(y-5)^2&=25 \end{array}$

ok i solved this by a lot of steps and got (1,1) as the intersection of all 3 circles
these has got to be other options to this.
basically I expanded the equations then set them equal to each other but what a mess:unsure:

suggestions?
I was thinking about a matrix but not sure how to set it up

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You can use this concept from the family of circles that $S_1 =0$ and $S_2 =0$ are two given circles intersecting each other then $S_1 - S_2 =0$ gives the equation of the line passing through those points.
For this question find any two lines using this and find their intersection point which will be your answer (here all the lines will be tangents)

There is a correction in my above statement I did not look at the figure carefully and typed about how to find the radical centre and equation of tangent which is wrong because one of the two circles are intersecting so... The correct one is --->>

Solving $S_1 - S_2=0$ will give the equation of chord joining the intersecting points... Find any two such chords then you have two equations and two unknowns (x,y).

just curious could the point of intersection of 3 circles be a limit

we are basically solving by deduction

what do you mean by limit?

basically are we not approaching a point?

I don't think so... We use limits when it is difficult to approach but here it is very easy to locate that point.

so basic algebra is the only way it can be done

karush said:
so basic algebra is the only way it can be done
I see the limit procedure as more of a numerical approach. We can, of course, try to solve it numerically. The algebraic solution is much better to my mind.

-Dan