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In this I took z = x + iy

Then solving the given equation , I got

$15x^2-40x+25=4x^2y^2-y^2$

Now how to proceed

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Your equation is incorrect. If $z=x+iy$, the equation should be : $$ x^2 + y^2 +4x-5=0 $$ This represents a circle with centre $(-2,0)$ and radius $3$. Adding $3+2i$, we get a circle with centre $1+2i$ and radius $3$.

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    How you have added 3+2i – 2017-02-19
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    With reference to the centre, all the points of the circle will lie at a distance. So to add $3+2i$ to the circle, so that the entire circle is moved to a new location. – 2017-02-19
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Hint: write it as:

$$ z \bar z + 2 z + 2 \bar z + 4 - 4 - 5 =0 \;\;\iff\;\; (z+2)(\bar z + 2) = 9 \;\;\iff\;\; \mid z+2\mid^2 = 9 $$

[ EDIT ]  The latter equation can be written in terms of $\,z+3+2i\,$ as:

$$ \mid z+2\mid = 3 \;\;\iff\;\; \mid z+3+2i - 3 - 2i+2\mid = 3 \;\;\iff\;\; \mid (z+3+2i) - (1+ 2i)\mid = 3 $$

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    Now in this how we will add 3+2i – 2017-02-19
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    @user123733 You don't add anything. In general $|z-c|=r$ is the locus of points $z$ at distance $r$ from center $c\,$ i.e. the circle of radius $r$ centered at $c\,$. The above reduced your equation to $|z+2|=3$ so the locus is the circle of radius $r=3$ and center $c=-2\,$. – 2017-02-19
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    But we have to find thelocus of z+3+2i – 2017-02-19
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    @user123733 Sorry, missed that part. Inlined at the end of the answer. – 2017-02-19