Question

Find the centre and the radius of the circle x²+y²+8x+10y-8=0​

Answer 1

Answer:

Step-by-step explanation:

Given, equation of the circle,

$\sf{x^2+y^2+8x+10y-8=0}$

$\tt{Centre\equiv(-4,-5)}$

$\tt{Radius=\sqrt{16+25+8}=\sqrt{49}=7}$

Answer 2

$\sf \: Given \: that \: the \: equation \: of \: a \: circle \\ \\ \small{\orange{ \boxed{\boxed{\begin{array}{cc} \rm \: {x}^{2} + {y}^{2} + 8x + 10y - 8 = 0\\ \\ \small{ \rm \implies \: {x}^{2} + {y}^{2} + 2 \times 4\times x + 2 \times 5 \times y - 8 = 0} \\ \\ \sf \: standard \: equation \: o f \: circle \: is : \\ \\ \rm \: a{x}^{2} + a{y}^{2} + 2gx + 2fy + c = 0 \\ \\ \sf \: by \: comparing \\ \\ \rm \: 2gx = 2 \times 4 \times x \implies g = 4 \\ \\ \rm \: 2fy = 2 \times 5 \times y \implies \: f = 5\\ \\ c = - 8 \\ \\\sf\:center\:(-g,-f)\:=\:(-4,-5) \\\\\sf \: radius \: \: r = \sqrt{ {g}^{2} + {f}^{2} - c } \\ \\ = \sqrt{ {( 4)}^{2} + ( { 5)}^{2} - ( - 8)} \\ \\ = \sqrt{16 + 25 + 8} \\ \\ = \sqrt{49} \\ \\ = 7 \: unit \\ \\ \\ \therefore \sf \: radius \: \: r = 7 \: unit\end{array}}}}}$