A circle has radius sqrt(10)/2 and pass through A(1,1).A line y = 3x - 7 pass through the centre of the circle.Find the equation of the circle?

Question

Here is my attempt

h = 3k -7 ----(1)

(h-1)^2 + (k -1)^2 = 10/4

(h-1)^2 + (3h - 8)^2 = 10/4

This second one doesn't working.Is my approch wrong?

P.S: Sorry for the typo.Also I assumed the center is C(h,k)

Answer 1

Given that the centre lies on the line $y=3x-7$ we can parameterise the centre to be the point $(t,3t-7)$. So the equation of the circle is given by $$(x-t)^2+(y-3t+7)^2=\frac52$$ for some $t\in\mathbb{R}$. As the point $A(1,1)$ lies on the circle, the equation must hold for $x=y=1$ hence $$(1-t)^2+(8-3t)^2=\frac52$$ This quadratic has a single repeated root namely $$t=\frac52$$ Hence the equation of the circle is $$\left(x-\frac52\right)^2+\left(y-\frac12\right)^2=\frac52$$

Answer 2

We want to find the point a distance of $\sqrt{10}/2$ on the line we are given from the point we are given. $$(x-1)^2+(3x-8)^2=5/2$$ $$10x^2-50x+65=5/2$$ $$x^2-5x+6.25=0$$ The point is therefore $$x=5/2$$ $$y=1/2$$ This is the only such point, since the equation only has one root, therefore it is the center of the circle. Thus the equation of the circle is $$(x-5/2)^2+(y-1/2)^2=5/2$$