Points of Intersection of Two Circles – Examples with Answers

Step 1: By subtracting the equations of the circles, we can find the equation of the common chord:

$latex x^2+y^2-3x+5y-4=0$

$latex x^2+y^2-x+4y-7=0~(-$

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$latex -2x+y+3=0$

Step 2: Solving the equation for y, we have:

$latex y=2x-3$

Step 3: Substituting the equation from step 2 into the equation of the second circle, we have:

$latex x^2+y^2-x+4y-7=0$

$$x^2+(2x-3)^2-x+4(2x-3)-7=0$$

$latex 5x^2-5x-10=0$

$latex 5(x-2)(x+1)=0$

Solving, we have $latex x=2$ and $latex x=-1$.

Step 4: When $latex x=2$, we have $latex y=1$ and when $latex x=-1$, we have $latex y=-5$.

Then, the intersection points of the circles are $latex (-1,~-5)$ and $latex (2, 1)$.

Step 1: By subtracting the equations of the circles, we have:

$latex x^2+y^2-5x+3y-4=0$

$latex x^2+y^2-4x+6y-12=0~(-$

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$latex -x-3y+8=0$

Step 2: Solving the equation for x, we have:

$latex x=-3y+8$

Step 3: Substituting the equation from step 2 into the equation of the first circle, we have:

$latex x^2+y^2-5x+3y-4=0$

$$(-3y+8)^2+y^2-5(-3y+8)+3y-4=0$$

$latex 10y^2-30y+20=0$

$latex 10(y-2)(y-1)=0$

Solving, we have $latex y=2$ and $latex y=1$.

Step 4: When $latex y=2$, we have $latex x=2$ and when $latex y=1$, we have $latex x=5$.

Then, the intersection points of the circles are $latex (2, ~2)$ and $latex (5,~ 1)$.

Step 1: Let’s find the equation of the common chord by subtracting the equations of the circles:

$latex x^2+y^2-4x+3y+5=0$

$latex x^2+y^2-6x+5y+9=0~(-$

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$latex 2x-2y-4=0$

Step 2: We can simplify the equation obtained by dividing by 2, and solving for x, we have:

$latex x-y-2=0$

$latex x=y-2$

Step 3: Let’s substitute the equation from step 2 into the equation of the first circle. Then, we have:

$latex x^2+y^2-4x+3y+5=0$

$$(y-2)^2+y^2+4(y-2)+3y+5=0$$

$latex 2y^2+3y+1=0$

$latex (2y+1)(y+1)=0$

Solving, we have $latex y=-\frac{1}{2}$ and $latex y=-1$.

Step 4: When $latex y=-\frac{1}{2}$, we have $latex x=1~\frac{1}{2}$ and when $latex y=-1$, we have $latex x=1$.

Therefore, the points of intersection of the circles are $latex (1 ~\frac{1}{2},~-\frac{1}{2})$ and $latex (1, -1)$.

Step 1: Subtracting the equations of the circles, we have:

$latex x^2+y^2+3x-2y-7=0$

$latex x^2+y^2+x-y-8=0~(-$

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$latex 2x-y+1=0$

Step 2: Solving the equation for y, we have:

$latex y=2x+1$

Step 3: Using the equation from step 2 in the equation of the second circle, we have:

$latex x^2+y^2+x-y-8=0$

$$x^2+(2x+1)^2+x-(2x+1)-8=0$$

$latex 5x^2+3x-8=0$

$latex (5x+8)(x-1)=0$

Solving, we have $latex x=-\frac{8}{3}$ and $latex x=1$.

Step 4: When $latex x=-\frac{8}{3}$, we have $latex y=-\frac{11}{5}$ and when $latex x=1$, we have $latex y=3$.

Therefore, the points of intersection of the circles are $latex (-\frac{8}{3},~\frac{11}{5})$ and $latex (1, 3)$.

Step 1: We find the equation of the common chord by subtracting the equations of the circles:

$latex x^2+y^2-3x+13y-48=0$

$latex x^2+y^2+x-3y=0~~(-$

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$latex -4x+16y-48=0$

Step 2: We can simplify the equation obtained by dividing by 4, and solving for x, we have:

$latex -x+4y-12=0$

$latex x=4y-12$

Step 3: Substituting the equation from step 2 into the equation of the second circle, we have:

$latex x^2+y^2+x-3y=0$

$$(4y-12)^2+y^2+(4y-12)-3y=0$$

$latex 17y^2-95y+132=0$

$latex (17y-44)(y-3)=0$

Solving, we have $latex y=\frac{44}{17}$ and $latex y=3$.

Step 4: When $latex y=\frac{44}{17}$, we have $latex x=-\frac{28}{17}$ and when $latex y=3$, we have $latex x=0$.

Therefore, the points of intersection of the circles are $latex (-\frac{28}{17},~\frac{44}{17})$ and $latex (0, 3)$.