+0  
 

Best Answer 

 #1
avatar+18629 
+2

What is the general form of the equation of a circle with its center at (-2, 1) and passing through (-4, 1)?

 

center at \((x_c=-2,\ y_c= 1 )\)

passing through point at \((x_p=-4,\ y_p= 1 )\)

\(\text{radius}^2 = r^2 = (x_c-x_p)^2+(y_c-y_p)^2\)

 

general form of the equation of a circle: \((x-x_c)^2 +(y-y_c)^2 = r^2\)

 

so we have:

\(\small{ \begin{array}{|rcll|} \hline (x-x_c)^2 +(y-y_c)^2 &=& r^2 \qquad | \quad r^2 = (x_c-x_p)^2+(y_c-y_p)^2 \\ (x-x_c)^2 +(y-y_c)^2 &=& (x_c-x_p)^2+(y_c-y_p)^2 \\ x^2-2x_c\cdot x + \not{x_c^2} + y^2-2y_c\cdot y + \not{y_c^2} &=& \not{x_c^2} -2x_cx_p + x_p^2 + \not{y_c^2}-2y_cy_p + y_p^2\\ x^2-2x_c\cdot x + y^2-2y_c\cdot y &=& -2x_cx_p + x_p^2 + -2y_cy_p + y_p^2 \\ x^2-2x_c\cdot x + y^2-2y_c\cdot y + 2x_cx_p - x_p^2 + 2y_cy_p - y_p^2 &=& 0 \\ x^2+ y^2 -2x_c\cdot x -2y_c\cdot y + 2x_cx_p + 2y_cy_p - x_p^2 - y_p^2 &=& 0 \\ x^2+ y^2 -2x_c\cdot x -2y_c\cdot y + x_p\cdot(2x_c-x_p) + y_p\cdot(2y_c-y_p) &=& 0 \\ \hline \end{array} } \)

 

The general form of the equation of a circle with its center \((x_c,y_c) \)and passing through point \( (x_p,y_p) \) is:

\(\mathbf{x^2+ y^2 -2x_c\cdot x -2y_c\cdot y + x_p\cdot(2x_c-x_p) + y_p\cdot(2y_c-y_p) = 0} \)

 

\(\small{ \begin{array}{|lrcll|} \hline x_c=-2,\ y_c= 1 \\ x_p=-4,\ y_p= 1 \\\\ & x^2+ y^2 -2x_c\cdot x -2y_c\cdot y + x_p\cdot(2x_c-x_p) + y_p\cdot(2y_c-y_p) &=& 0 \\ & x^2+ y^2 -2\cdot (-2)\cdot x -2\cdot 1\cdot y + (-4)\cdot[2\cdot(-2)-(-4)] + 1\cdot(2\cdot 1-1) &=& 0 \\ & x^2+ y^2 +4x -2y + (-4)\cdot(-4+4) + 1\cdot(2-1) &=& 0 \\ & x^2+ y^2 +4x -2y + 0 + 1\cdot 1 &=& 0 \\ & x^2+ y^2 +4x -2y + 1 &=& 0 \\ \hline \end{array} }\)

 

The equation of the circle is \(x^2+ y^2 +4x -2y + 1 = 0 \)

 

laugh

heureka  Jun 7, 2017
edited by heureka  Jun 7, 2017
Sort: 

1+0 Answers

 #1
avatar+18629 
+2
Best Answer

What is the general form of the equation of a circle with its center at (-2, 1) and passing through (-4, 1)?

 

center at \((x_c=-2,\ y_c= 1 )\)

passing through point at \((x_p=-4,\ y_p= 1 )\)

\(\text{radius}^2 = r^2 = (x_c-x_p)^2+(y_c-y_p)^2\)

 

general form of the equation of a circle: \((x-x_c)^2 +(y-y_c)^2 = r^2\)

 

so we have:

\(\small{ \begin{array}{|rcll|} \hline (x-x_c)^2 +(y-y_c)^2 &=& r^2 \qquad | \quad r^2 = (x_c-x_p)^2+(y_c-y_p)^2 \\ (x-x_c)^2 +(y-y_c)^2 &=& (x_c-x_p)^2+(y_c-y_p)^2 \\ x^2-2x_c\cdot x + \not{x_c^2} + y^2-2y_c\cdot y + \not{y_c^2} &=& \not{x_c^2} -2x_cx_p + x_p^2 + \not{y_c^2}-2y_cy_p + y_p^2\\ x^2-2x_c\cdot x + y^2-2y_c\cdot y &=& -2x_cx_p + x_p^2 + -2y_cy_p + y_p^2 \\ x^2-2x_c\cdot x + y^2-2y_c\cdot y + 2x_cx_p - x_p^2 + 2y_cy_p - y_p^2 &=& 0 \\ x^2+ y^2 -2x_c\cdot x -2y_c\cdot y + 2x_cx_p + 2y_cy_p - x_p^2 - y_p^2 &=& 0 \\ x^2+ y^2 -2x_c\cdot x -2y_c\cdot y + x_p\cdot(2x_c-x_p) + y_p\cdot(2y_c-y_p) &=& 0 \\ \hline \end{array} } \)

 

The general form of the equation of a circle with its center \((x_c,y_c) \)and passing through point \( (x_p,y_p) \) is:

\(\mathbf{x^2+ y^2 -2x_c\cdot x -2y_c\cdot y + x_p\cdot(2x_c-x_p) + y_p\cdot(2y_c-y_p) = 0} \)

 

\(\small{ \begin{array}{|lrcll|} \hline x_c=-2,\ y_c= 1 \\ x_p=-4,\ y_p= 1 \\\\ & x^2+ y^2 -2x_c\cdot x -2y_c\cdot y + x_p\cdot(2x_c-x_p) + y_p\cdot(2y_c-y_p) &=& 0 \\ & x^2+ y^2 -2\cdot (-2)\cdot x -2\cdot 1\cdot y + (-4)\cdot[2\cdot(-2)-(-4)] + 1\cdot(2\cdot 1-1) &=& 0 \\ & x^2+ y^2 +4x -2y + (-4)\cdot(-4+4) + 1\cdot(2-1) &=& 0 \\ & x^2+ y^2 +4x -2y + 0 + 1\cdot 1 &=& 0 \\ & x^2+ y^2 +4x -2y + 1 &=& 0 \\ \hline \end{array} }\)

 

The equation of the circle is \(x^2+ y^2 +4x -2y + 1 = 0 \)

 

laugh

heureka  Jun 7, 2017
edited by heureka  Jun 7, 2017

25 Online Users

avatar
avatar
avatar
avatar
avatar
avatar
avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details