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# What is the general form of the equation of a circle with its center at (-2, 1) and passing through (-4, 1)? x2 + y2 − 4x + 2y + 1 = 0

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see title of question

Guest Jun 7, 2017

#1
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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 $$x^2+ y^2 +4x -2y + 1 = 0$$

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

#1
+19197
+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 $$x^2+ y^2 +4x -2y + 1 = 0$$

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

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