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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 ?$

Short Answer

Expert verified
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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