If the equation `p x^2+(2-q)x y+3y^2-6q x+30 y+6q=0`
represents a circle, then find the values of `p` and `q`.
Text Solution
Verified by Experts
In the equation of circle, the xy term does not occur and the coefficients of `x^(2)` and `y^(2)` are equal. Therefore, `2-q=0` or `q=2` and `p=3` Also, for these values of p and q, the circle is real.
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