### Instruction: Find the equation of the circle which passes through $$(2, 3)$$ and $$(4, 5)$$ and the centre lies on the straight line $$y-4x+3=0$$.### Response: To find the equation of the circle, I need to know its center and radius.
The center of the circle lies on the given line, so I can write its coordinates as $$(x, 4x-3)$$ for some value of $$x$$.
The radius of the circle is the distance from the center to any of the given points, say $$(2, 3)$$.
Using the distance formula, the radius squared is equal to $$(x-2)^2+(4x-3-3)^2 = (x-2)^2+(4x-6)^2$$.
I can simplify this expression by expanding the squares and collecting like terms: $$r^2 = x^2-4x+4+16x^2-48x+36 = 17x^2-52x+40$$.
Now I have an equation relating $$x$$ and $$r$$, but I want to eliminate $$r$$ and find $$x$$.
To do that, I can use the fact that the circle also passes through $$(4, 5)$$, which means that the radius squared is also equal to $$(x-4)^2+(4x-3-5)^2 = (x-4)^2+(4x-8)^2$$
Expanding and simplifying this expression gives me another equation: $$r^2 = x^2-8x+16+16x^2-64x+64 = 17x^2-72x+80$$
Since both expressions are equal to $$r^2$$, I can set them equal to each other and solve for $$x$$: $$17x^2-52x+40=17x^2-72x+80$$
Subtracting $$17x^2$$ from both sides and rearranging, I get $$20x = 40$$, which implies $$x=2$$
Now that I have $$x$$, I can plug it into the coordinates of the center to get $$(2, 4(2)-3) = (2, 5)$$
I can also plug $$x=2$$ into the expression for the radius squared to get $$r^2 = 17(2)^2-52(2)+40 = 16$$
Therefore, the radius is $$r=4$$
Finally, I can write the equation of the circle using the standard form: $$(x-h)^2+(y-k)^2=r^2$$, where $$(h, k)$$ is the center and $$r$$ is the radius.