Explanation:
A Given,
\(x^2+y^2-5 x+6 y+15=0\)
\(x^2+y^2-2 x+6 y+6=0\)
From equation (i), we get -
\(x^2+y^2-5 x+6 y+15=0\)
\(2 \mathrm{~g}_1=-5,2 \mathrm{f}_1=6, \mathrm{C}_1=15\)
\(\mathrm{~g}_1=\frac{-5}{2}, \mathrm{f}_1=3, \mathrm{C}_1=15,\left(-\mathrm{g}_1,-\mathrm{f}_1\right) \rightarrow \mathrm{C}_1=\left(\frac{5}{2},-3\right)\)
\(\mathrm{S}_1=\sqrt{\mathrm{g}_1{ }^2+\mathrm{f}_1^2-\mathrm{C}_1}=\sqrt{\left(\frac{-5}{2}\right)^2+3^2-15}=\sqrt{\frac{25}{4}+9-15}\)
\(=\sqrt{\frac{25}{4}-6}=\sqrt{\frac{1}{4}}=\frac{1}{2}\)
From equation (ii), we get -
\(x^2+y^2-2 x+6 y+6=0\)
\(2 g_2=-2,2 f_2=6, C_2=6\)
\(g_2=-1, f_2=3, C_2=6, C_2 \rightarrow\left(-g_2, f_2\right)=(1,-3)\)
\(\mathrm{S}_2=\sqrt{\mathrm{g}_2^2+\mathrm{f}_2^2-\mathrm{C}_2}=\sqrt{(-1)^2+(3)^2-6}=\sqrt{1+9-6}=\sqrt{4}=2\)
\(\therefore \quad \mathrm{C}_1 \mathrm{C}_2=\sqrt{\left(\frac{5}{2}-1\right)^2+(-3-(-3))^2}=\sqrt{\frac{9}{4}+0}\)
\(\mathrm{C}_1 \mathrm{C}_2=\frac{3}{2}\)
\(\therefore \quad \mathrm{C}_1 \mathrm{C}_2=\left|\mathrm{S}_1-\mathrm{S}_2\right|\)
\(\frac{3}{2}=\left|\frac{1}{2}-2\right|=\frac{3}{2}\)
since
Hence, the two circle touch each other internally.