Question

Write the equation of the circle passing through (3,4) and touching y-axis at the origin.

Answer 1

Circle is touching y-axis at the origin.
Thus, centre of circle is on x-axis. So
Let centre of circle is (h, 0)
Equation of circle is\\
$(x-h{)}^2(y-0{)}^2=r2\dots \left(1\right)$
It is passing through (3, 4)
$(3-h{)}^2+(4{)}^2=r^2(3-h{)}^2+16=r^2\dots (2)$
It is also passing through (3, 4)
$(0-h{)}^2+(0-0{)}^2=r^2{h}^2=r^2$
Now, equation (2) becomes,
$(3-h{)}^2+16=h^{2}9-6h+h^2+16-6h+25=0h=\frac{25}{6}$
Equation of circle is,
$(x-h{)}^2+y^2=h^2{x}^2-2xh+y^2=0x^2-2xh+y^2=0x^2-2x\left(\frac{25}{6}\right)+y^2=06x^2-50x+6y^2=03x^2-25x+3y^2=03(x^2+y^2)-25x=0$