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Question Number 187765 by Tawa11 last updated on 21/Feb/23

Answered by cortano12 last updated on 21/Feb/23

$$\left(1\mathrm{i}\right)\Rightarrow (\mathrm{x}-1)(\mathrm{x}-3)+(\mathrm{y}-3)(\mathrm{y}-7)=0$$$$\Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{x}-10\mathrm{y}+24=0$$

Answered by cortano12 last updated on 21/Feb/23

$$(1.2)\mathrm{r}=\mid 4\mid =4$$$$\Rightarrow {(\mathrm{x}-3)}^2+{(\mathrm{y}-4)}^2=16$$$$\Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-6\mathrm{x}-8\mathrm{y}+9=0$$

Answered by floor(10²Eta[1]) last updated on 21/Feb/23

$$\left(1\right):\mathrm{AB}=2ı+4ȷ\Rightarrow \parallel \mathrm{AB}\parallel =\sqrt{4+16}=\sqrt{20}=2\sqrt{5}$$$$\Rightarrow \mathrm{radius}=\sqrt{5}$$$$\mathrm{Center}:(\frac{3+1}{2},\frac{7+3}{2})=(2,5)$$$$\Rightarrow {(\mathrm{x}-2)}^2+{(\mathrm{y}-5)}^2=5$$$$$$$$\left(2\right):\mathrm{r}=4$$$${(\mathrm{x}-3)}^2+{(\mathrm{y}-4)}^2=16$$

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