Answer:
(3,4)
Step-by-step explanation:
[tex]3x+4y=25[/tex]
[tex]x^2+y^2=25[/tex]
From the first equation
[tex]x=\frac{25-4y}{3}[/tex]
Applying to the second equation
[tex]\left(\frac{25-4y}{3}\right)^2+y^2=0\\\Rightarrow \frac{625+16y^2-200y+9y^2}{9}=25\\\Rightarrow 625+16y^2-200y+9y^2=9\times 25\\\Rightarrow 25y^2-200y+625=225\\\Rightarrow 25y^2-200y+400=0\\\Rightarrow y^2-8y+16=0[/tex]
[tex]y_{1,\:2}=\frac{-\left(-8\right)\pm \sqrt{\left(-8\right)^2-4\cdot \:1\cdot \:16}}{2\cdot \:1}\\\Rightarrow y=4[/tex]
y=4.
Point on the circle
[tex]x=\sqrt{25-16}=3[/tex]
So, the line will intersect at the point (3,4)
