Hayley Steele

2023-02-26

Two congruent circles with centres at (2,3) and (5,6), which intersect at right angles, have radius equal to?

Cailyn Knight

Beginner2023-02-27Added 7 answers

Circles naturally cross at a 90 degree angle.

Given that circles have the same radius, they are congruent.

Then ${r}_{1}={r}_{2}=r$

${C}_{1}{C}_{2}=\sqrt{(5-2{)}^{2}+(6-3{)}^{2}}\phantom{\rule{0ex}{0ex}}{C}_{1}{C}_{2}=\sqrt{18}$

By pythagoros theorem

${r}^{2}+{r}^{2}=({C}_{1}{C}_{2}{)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 2{r}^{2}=(\sqrt{18}{)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {r}^{2}=9\phantom{\rule{0ex}{0ex}}\Rightarrow r=3$

Given that circles have the same radius, they are congruent.

Then ${r}_{1}={r}_{2}=r$

${C}_{1}{C}_{2}=\sqrt{(5-2{)}^{2}+(6-3{)}^{2}}\phantom{\rule{0ex}{0ex}}{C}_{1}{C}_{2}=\sqrt{18}$

By pythagoros theorem

${r}^{2}+{r}^{2}=({C}_{1}{C}_{2}{)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 2{r}^{2}=(\sqrt{18}{)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {r}^{2}=9\phantom{\rule{0ex}{0ex}}\Rightarrow r=3$