In the co-axial system of circle {x^2} + {y^2 | vedclass

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Question

(b) Given, equation of the circle $x^2 + y^2 + 2gx + c = 0$

where $c$ is constant and $g$ represents the parameter of a coaxial system and $c >

Vedclass · Accepted Answer

Touching type

Vedclass · Answer

(b) Given, equation of the circle $x^2 + y^2 + 2gx + c = 0$

where $c$ is constant and $g$ represents the parameter of a coaxial system and $c > 0.$

We know that the standard equation of a circle is ${x^2} + {y^2} + 2gx + 2fy + c = 0.$

Comparing the given equation with the standard equation, we get centre $ \equiv ( - g,\,0)$ and radius $\sqrt {{g^2} - c} $.

Therefore radius becomes zero, when ${g^2} - c = 0$ or $g = \pm \sqrt c .$

Therefore $(\sqrt c ,\,0)$ and $( - \sqrt c ,\,0)$ are the limiting points of the coaxial system of circles.

Since $c > 0$, therefore $\sqrt c $  is real and limiting points are real and distinct.

Thus the co-axial system is said to be touching type.

In the co-axial system of circle ${x^2} + {y^2} + 2gx + c = 0$, where $g$ is a parameter, if $c > 0$ then the circles are

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