Manipal Entrance Test 2020 Math Paper

Given system of circles is $x^{2}+y^{2}+4 x+7=0$.....(i) $2(x^{2}+y^{2})+3 x+5 y+9=0$ or $\;\; x^{2}+y^{2}+\;\frac{3}{2} x+\;\frac{5}{2} y+\;\frac{9}{2}=0$ .....(ii) and $x^{2}+y^{2}+y=0$......(iii) The radical centre can be obtained by solving the Eqs. (i), (ii) and (iii). On subtracting Eq. (ii) from Eq. (i), we get $4 x-\;\frac{3}{2} x-\;\frac{5}{2} y+7-\;\frac{9}{2}=0$ $\Rightarrow \;\; \;\frac{5}{2} x-\;\frac{5}{2} y+\;\frac{5}{2}=0$ $\Rightarrow \;\; x-y+1=0$....(iv) On subtracting Eq. (iii) from Eq. (i), we get $4 x-y+7=0$...(v) On solving Eqs. (iv) and (v), we get $x=-2$ and $y=-1$ Hence, radical centre is $(-2,-1)$.