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$C_m$ is a family of circles ($m\in \mathbb{R}$):

$C_{m}: x^2+y^2-2mx+(m+2)y-3m-4=0$

  1. Determine the set of centers $(D)$ when $m$ changes in $\mathbb{R}$

  2. Prove that all circles in $C_m$ pass through two fixed points $A$ and $B$, and prove that $(AB)\perp(D)$

  3. Determine depending on values of $m$ the number of circles passing through the point $M_0(x_0,y_0) $

Completing the square we get: $C_{m}: (x-m)^2+(y+\frac{m+2}2)^2 = \frac{5}{4}m^2+4m+5$

Thus, the center is $\Omega(m,-\frac{m+2}{2})$ and radius is $r=\sqrt{\frac{5}{4}m^2+4m+5}$

For the first question:

The center of the circles is $\Omega(m,-\frac{m+2}{2})$

$y=-\frac{x+2}{2} \implies (D): x+2y+2=0$

For question (2) and (3), I'm not sure from where to start.

Here a graph: here

Bilal HIMITE
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  • Hint for part 3: three distinct noncolinear points define a unique circle. Are there any values of $m$ for which this condition doesn’t hold? – amd Jan 02 '18 at 00:18

1 Answers1

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For the second question, the given equation $$x^2+y^2+2y-4+m(-2x+y-3)=0$$ represents all arbitrary circle passing through the intersection of $$x^2+y^2+2y-4=0$$

and $$-2x+y-3=0$$

So, the intersections are the two fixed points.

lab bhattacharjee
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