Question

Find the equation of the circle with radius 5 whose centre lies on x - axis and passes through the point (2,3)

Answer 1

Since the centre of he circle lies on x - axis, the coordinates of centre is (h,O).
now the circle passes through the point (2,3).
$\therefore \text{Radius of circle}=\sqrt{(h-2{)}^{+}(0-3{)}^2}=\sqrt{h^2+4-4h+9}=\sqrt{h^2-4h+13}$
But radius of cirlcle = 5
$\therefore \sqrt{h^2-4h+13}=5\Rightarrow h^2-4h+13=25\Rightarrow h^2-4h-12=0\Rightarrow (h-6)(h+2)=0\Rightarrow h=6orh=-2\text{When h = 6}\text{Equation of required circle is}(x-6{)}^2+(y-0{)}^2=(5{)}^2\Rightarrow x^2+36-12x+y^2=25\Rightarrow x^2+y^2-12x+11=0\text{When h = - 2}\text{Equation of required circle is}(x+2{)}^2+(y-0{)}^2=(5{)}^2\Rightarrow x^2+4+4x+y^2=25\Rightarrow x^2+y^2+4x-21=0$