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Problem 14

# For what values of $$l$$ and $$m$$, circle $$5\left(x^{2}+y^{2}\right)+b y-m=0$$ belongs to the co-axial system determined by the circles $x^{2}+y^{2}+2 x+4 y-6=0$$and$$2\left(x^{2}+y^{2}\right)-x=0 ?$

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The values of $$l$$ and $$m$$ that make the given circle belong to the coaxial system determined by the two provided circles are $$2$$ and approximately $$1.06$$ respectively.
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## Step 1: Determine the coefficients of the two circles

The equation of the first provided circle can be rewritten as $$x^{2}+y^{2}+2(-1)x+2(2)y-6=0$$, thus $$g_1=-1$$, $$f_1=2$$, and $$c_1=-6$$. Similarly for the second circle, we rewrite it as $$2x^{2}+2y^{2}-2(0.5)x+0y+0=0$$, hence $$g_2=0.5$$, $$f_2=0$$, and $$c_2=0$$. We can therefore conclude that the equation for any circle within this coaxial system will have the form $$5\left(x^{2}+y^{2}\right)+10ax+10by+c=0.$$

## Step 2: Determine the center and radius of the auxiliary circle

The center $$-g, -f$$ of the auxiliary circle for this system can be obtained by averaging the centers of the two provided circles giving us $$-g=(g_1+g_2)/2$$ which equals $$-0.25$$ and $$-f=(f_1+f_2)/2$$ which equals $$1$$. The radius $$a$$ of the auxiliary circle is given by $$\sqrt{g^2+f^2-c}$$, but since $$c=0$$ for the given circles, the radius $$a$$ is simply $$\sqrt{g^2+f^2} = \sqrt{(0.25)^2+1^2} = 1.03$$ to three significant figures. Therefore, the auxiliary or Soddy's circle for the given coaxial system is $$x^2+y^2+0.5x-2y=0$$.

## Step 3: Determine the values for $$l$$ and $$m$$

The required circle must be in the form $$a\left(x^{2}+y^{2}\right)+2gx+2fy+c=0$$. Comparing this with the given expression for the circle, it can be concluded that the equation for $$b$$ will be $$b=2f$$, hence $$b=2(1)=2$$. Similarly, $$m=a^2=1.03^2=1.06$$. Therefore, the values of $$l$$ and $$m$$ are $$2$$ and $$1.06$$ respectively.

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