The two circles {x^2} + {y^2} - 2x + 6y + 6 = | vedclass

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Question

(c) Given, equations of the circles

${x^2} + {y^2} - 2x + 6y + 6$$=0$ .....$(i)$

and ${x^2} + {y^2} - 5x + 6y + 15 = 0$ .....$(ii)$

We know t

Vedclass · Accepted Answer

Touch internally

Vedclass · Answer

(c) Given, equations of the circles

${x^2} + {y^2} - 2x + 6y + 6$$=0$ .....$(i)$

and ${x^2} + {y^2} - 5x + 6y + 15 = 0$ .....$(ii)$

We know that the standard equation of a circle is ${x^2} + {y^2} + 2gx + 2fy + c = 0.$

Therefore for circle $(i),$ $g = - 1;\,f = 3;\,\,\,c = 6;$ centre $A = (1,\, - 3)$

and radius $({r_1}) = \sqrt {{g^2} + {f^2} - c} = \sqrt {1 + 9 - 6} = 2$.

Similarly, for circle $(ii),$ $g = \frac{{ - 5}}{2};\,\,f = 3;\,c = 15;$

 Centre $B \equiv \,\left( { + \frac{5}{2}, - 3} \right)$

and radius $({r_2}) = \sqrt {\frac{{25}}{4} + 9 - 15} = \frac{1}{2}$

Therefore distance between $A$ and $B$ $ = \sqrt {{{\left( {\frac{5}{2} - 1} \right)}^2} + {{( - 3 + 3)}^2}} = \frac{3}{2}$

and difference of radii $({r_1} - {r_2}) = 2 - \frac{1}{2} = \frac{3}{2}.$

Since distance between $A$ and $B$ is equal to ${r_1} - {r_2},$

therefore the circles touch each other internally.

The two circles ${x^2} + {y^2} - 2x + 6y + 6 = 0$ and ${x^2} + {y^2} - 5x + 6y + 15 = 0$

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