Two circles x^2 + y^2 = ax and x^2 + y^2 = c^ | vedclass

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(A) The given equations are $x^2 + y^2 - ax = 0$ and $x^2 + y^2 = c^2$.For the first circle,the centre $C_1 = (\frac{a}{2}, 0)$ and radius $r_1 = |\fr

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$|a| = c$

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(A) The given equations are $x^2 + y^2 - ax = 0$ and $x^2 + y^2 = c^2$.For the first circle,the centre $C_1 = (\frac{a}{2}, 0)$ and radius $r_1 = |\frac{a}{2}|$.For the second circle,the centre $C_2 = (0, 0)$ and radius $r_2 = |c|$.The distance between the centres is $d = \sqrt{(\frac{a}{2} - 0)^2 + (0 - 0)^2} = |\frac{a}{2}|$.Two circles touch each other if the distance between their centres is equal to the sum or difference of their radii,i.e.,$d = |r_1 \pm r_2|$.$|\frac{a}{2}| = ||\frac{a}{2}| \pm |c||$.Case $1$: $|\frac{a}{2}| = |\frac{a}{2}| + |c| \Rightarrow |c| = 0$ (not possible for a circle).Case $2$: $|\frac{a}{2}| = | |\frac{a}{2}| - |c| |$.This implies $|\frac{a}{2}| = |c| - |\frac{a}{2}|$ (assuming $|c| > |\frac{a}{2}|$) or $|\frac{a}{2}| = |\frac{a}{2}| - |c|$ (not possible).Thus,$2|\frac{a}{2}| = |c|$,which simplifies to $|a| = |c|$ or $|a| = c$ (since $c$ is a radius,$c > 0$).

Two circles $x^2 + y^2 = ax$ and $x^2 + y^2 = c^2$ touch each other if:

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