Find the equation of the circle which has its centre at \((-2 .\) 3) and passes through the point (4.5).
A. \(x^{2}+2 x+y^{2}-3 y\) \(-27=0\) B. \(x^{2}+y^{2}+4 x-6 y-27=0\) C. \(x^{2}+y^{2}-4 x+6 y\) \(+27=0\) D. \(x^{2}+y^{2}-2 x+3 y-27=0\)
Correct Answer: B
Explanation
let the equation of the circle be \((x-a)^{2}+(y-b)^{2}=r^{2}\) where \((a . b)\) is the centre and \(r\) is the radius. to find \(r\). we use distance between two points i.e. the distance between the centre \((-2.3)\) and the point \((4.5)\). \begin{array}{l} r=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \\ r=\sqrt{[4-(-2)]^{2}+(5-3)^{2}} \\ =\sqrt{(4+2)^{2}+(2)^{2}} \\ =\sqrt{6^{2}+7^{2}}=\sqrt{36+4} \\ r=\sqrt{40} \cdot r^{2}=40 \\ \Rightarrow(x-a)^{2}+(y-b)^{2}=r^{2} \\ {[x-(-2)]^{2}+(y-3)^{2}=40} \\ (x+2)^{2}+(1-3)^{2}=40 \\ x^{2}+4 x+4+y^{2}-6 y+9=40 \\ x^{2}+y^{2}+4 x-6 y+13-40=0 \\ x^{2}+y^{2}+4 x-6 y-27=0 \\ \end{array}