[tex]\displaystyle \underline{\text{circunfer{\^e}ncia}}:\\\\ (\text x-2)^2+(\text y+1)^2=4 \\\\ \text{centro}: \\\ \text x=2\ ; \ \text y = -1 \\\\\\ \underline{\text{retas}}: \\\\ 2\text x-3\text y+5=0\\ \text x-2\text y+4=0 \\[/tex]
Ponto P é a interseção das retas, então vamos resolver esse sistema :
[tex]2\text x-3\text y+5=0\\\\ \text x-2\text y+4=0 \to \boxed{\text x = 2\text y-4} \\\\ \underline{\text{substituindo na primeira reta}}: \\\\ 2(2\text y-4)-3\text y+5=0 \\\\ 4\text y-8-3\text y+5=0 \\\\ \text y = 8-5 \\\\ {\ \text{y}= 3\ } \\\\ \therefore \\\\ \text x = 2.3-4\\\\ \text x = 2[/tex]
Portanto o ponto P = ( 2, 3 )
Distância do ponto P até o centro da circunferência :
[tex]\text D = \sqrt{(2-2)^2+(3-(-1))^2} \\\\ \text D = \sqrt{0^2+(3+1)^2 } \\\\ \text D =\sqrt{4^2} \\\\ \huge\boxed{\text D = 4 \ }\checkmark[/tex]
