Question

lete the squares for each quadratic, list the center and radius, then graph each circlela and poin Achi always beling its translated center: ) x2 + 2x + y2 - 4y = 4 ) 2x2 + 2y2 + 3x - 5y = 2 ) x2 + y2 + 3x = 4 . x2 + y2 + 4x = 0 + y2 + 2mx - 2ny = 0 equati focus is th to t r2 (b) x² + y2 - 4x = 0 (d) x2 + y2 - 2x - 8y = 8 (f) 4x² + 4y2 - 16x + 24y = -27 (h) x2 + y2 - 7y = 0 (i) x² + y2 - 2ax + 2by = c termine which of the following equations represents a circle with a real non-zero radius: x2 + y2 + 10x = -30 (b) 3x2 + 3y2 - 11x = -91 4x2 + 4y2 + 18x - 8y = -85 (d) 36x2 + 36y2 - 36x + 48y =-16 th F

Answer 1

Let's begin by listing out the given information:

`\begin{gathered} 36x^2+36y^2-36x+48y=-16 \\ \text{dividing through by 36 to remove the coefficients of }x^2,y^2\text{, we have:} \\ (36)/(36)x^2-(36)/(36)x+(36)/(36)y^2+(48)/(36)y=-(16)/(36) \\ x^2-x+y^2+(4)/(3)y=-(4)/(9) \\ x^2-x+y^2+(4)/(3)y+(4)/(9)=0 \\ Comparing\text{ }this\text{ }with\text{ }the\text{ }general\text{ }equation\text{ of }circle\text{ }we\text{ have:} \\ x^2+y^2+2gx+2fy+c=0 \\ \Rightarrow2g=-1 \\ g=-(1)/(2) \\ 2f=(2)/(3) \\ \Rightarrow f=(1)/(3) \\ c=(4)/(9) \\ \Rightarrow c=(4)/(9) \end{gathered}`

Thus the center of the circle is:

`\begin{gathered} (h,k)=(-g,-f) \\ \Rightarrow(--(1)/(2),-(1)/(3)) \\ \Rightarrow((1)/(2),-(1)/(3)) \\ \\ r=\sqrt{g^(2)+f^(2)-c} \\ r=\sqrt[]{((1)/(2))^2+(-(1)/(3))^2-(4)/(9)} \\ r=\sqrt[]{(1)/(4)+(1)/(9)-(4)/(9)}=\sqrt[]{-(1)/(12)} \\ r=\sqrt[]{-(1)/(12)} \end{gathered}`