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Mengchun Zhang
Mengchun Zhang
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Here is the plot: circle problem where $A,B$ are the intersection of $O,O_1$, and $C,D$ are the intersection of $O,O_2$.

Since $AB=4$, segment $AB$ is the diameter of $O_1$, hence $O$ is on the normal bisector of $AB$, that is, $OO_1$ is perpendicular to $AB$.

Similarly, $OO_2$ is perpendicular to $CD$.

Therefore, $AO^2=(AO_1)^2+(OO_1)^2=4+(OO_1)^2$ and $CO^2=(CO_2)^2+(OO_2)^2=1+(OO_2)^2$. But since $AO=CO$, we have $$3+(OO_1)^2=(PO_2)^2.$$$$3+(OO_1)^2=(OO_2)^2.$$ Let $O=(x,y)$ then the above equation reads $$3+x^2+y^2=(5-x)^2+y^2,$$ which yields $x=\frac{11}5$, $y\in\mathbb R$.

Here is the plot: circle problem where $A,B$ are the intersection of $O,O_1$, and $C,D$ are the intersection of $O,O_2$.

Since $AB=4$, segment $AB$ is the diameter of $O_1$, hence $O$ is on the normal bisector of $AB$, that is, $OO_1$ is perpendicular to $AB$.

Similarly, $OO_2$ is perpendicular to $CD$.

Therefore, $AO^2=(AO_1)^2+(OO_1)^2=4+(OO_1)^2$ and $CO^2=(CO_2)^2+(OO_2)^2=1+(OO_2)^2$. But since $AO=CO$, we have $$3+(OO_1)^2=(PO_2)^2.$$ Let $O=(x,y)$ then the above equation reads $$3+x^2+y^2=(5-x)^2+y^2,$$ which yields $x=\frac{11}5$, $y\in\mathbb R$.

Here is the plot: circle problem where $A,B$ are the intersection of $O,O_1$, and $C,D$ are the intersection of $O,O_2$.

Since $AB=4$, segment $AB$ is the diameter of $O_1$, hence $O$ is on the normal bisector of $AB$, that is, $OO_1$ is perpendicular to $AB$.

Similarly, $OO_2$ is perpendicular to $CD$.

Therefore, $AO^2=(AO_1)^2+(OO_1)^2=4+(OO_1)^2$ and $CO^2=(CO_2)^2+(OO_2)^2=1+(OO_2)^2$. But since $AO=CO$, we have $$3+(OO_1)^2=(OO_2)^2.$$ Let $O=(x,y)$ then the above equation reads $$3+x^2+y^2=(5-x)^2+y^2,$$ which yields $x=\frac{11}5$, $y\in\mathbb R$.

Source Link
Mengchun Zhang
Mengchun Zhang
  • 3k
  • 12
  • 15

Here is the plot: circle problem where $A,B$ are the intersection of $O,O_1$, and $C,D$ are the intersection of $O,O_2$.

Since $AB=4$, segment $AB$ is the diameter of $O_1$, hence $O$ is on the normal bisector of $AB$, that is, $OO_1$ is perpendicular to $AB$.

Similarly, $OO_2$ is perpendicular to $CD$.

Therefore, $AO^2=(AO_1)^2+(OO_1)^2=4+(OO_1)^2$ and $CO^2=(CO_2)^2+(OO_2)^2=1+(OO_2)^2$. But since $AO=CO$, we have $$3+(OO_1)^2=(PO_2)^2.$$ Let $O=(x,y)$ then the above equation reads $$3+x^2+y^2=(5-x)^2+y^2,$$ which yields $x=\frac{11}5$, $y\in\mathbb R$.