Find the equation of family of circles which touch the pair of lines $x^{2} + y^{2}$ + 2y - 1 = 0

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Solution

The pair of lines is $x^{2} - (y - 1)^{2} = o$
or x + y 1 = 0 and x y + 1 = 0
if (h,k) be the center, then it will lie on the bisectors of these
$; ∴ \frac{h + k - 1}{\sqrt{2}} = \pm; \frac{h + k - 1}{\sqrt{2}} =; r$
$∴$ b either k = 1; or h = 0
When k = 1then h = $\pm \sqrt{2 r}$
when h = 0 then k = $1 \pm \sqrt{2 r}$
Hence the circles;are
$(x \pm \sqrt{2 r})^{2} + (y - 1)^{2} = r^{2}$ ;
or $(x - 0)^{2} + (y - 1 \mp; \sqrt{2})^{2} = r^{2}$
Above give the required families of circles, r being the parameter.
We call the above as family of\;circles because of parameter r. Here we are given only two conditions whereas a circle is completely determined if three conditions are given to find three unknowns g, f, c or h, k, r

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Find the equation of family of circles which touch the pair of lines x2+y2 + 2y - 1 = 0

Find the equation of family of circles which touch the pair of lines $x^{2} + y^{2}$ + 2y - 1 = 0