0
$\begingroup$

Find the equation of the circle which passes through the origin and cuts off intercepts $3$ and $4$ from the positive parts of the axes respectively.

My Attempt,

The circle cuts the axis of $X$ at $(3,0)$ and the axis of $Y$ at $(0,4)$. Let $r$ be the radius of the circle. Since the circle passes through the origin, $$r=\sqrt {h^2+k^2}$$,

Where, $(h,k)$ are the co ordinates of the centre of the circle.

2 Answers 2

2

Let the equation of the circle be $$x^2+y^2+2gx+2fy+c=0$$

Now it passes through $(0,0);(3,0);(0,4)$

We have three unknowns with there conditions, right?

  • 0
    As the circle passes through $(0,0)$ is it correct to consider $c=0$? – 2017-01-30
  • 0
    @NeWtoN, We don't need to consider. Put the values of $(x,y)$ one by one to find $g,f,c$ – 2017-01-30
  • 0
    @labbhattacharjee, $(x,y)=(0,0)$ does not imply $c=0$ from $$x^2 + y^2 + 2gx + 2fy + c = 0?$$ – 2017-01-30
  • 0
    @JoseArnaldoBebitaDris, Why? – 2017-01-30
  • 0
    @labbhattacharjee, you told the OP "We don't need to consider ($c=0$)". Would that assertion be correct? – 2017-01-30
  • 0
    @JoseArnaldoBebitaDris, My point was : we don't think it separately. – 2017-01-30
  • 0
    @lab bhattacharjee, could you please help me with this, too. math.stackexchange.com/questions/1937148/radius-4-circle-skewered-on-one-line-and-touching-another – 2017-01-30
  • 0
    @NeWtoN, An answer has been posted :) – 2017-01-30
2

Do it geometrically.

Bisect the line segment from $(0,0)$ to $(3,0)$. And the line segment from $(0,0)$ to $(0,4)$.

The center of the circle is obviously $(1.5,2)$.

The radius of the circle is $\sqrt {1.5^2 + 2^2}$.

The equation is thus: $$(x-1.5)^2+(y-2)^2=1.5^2 + 2^2$$

Simplify and expand to get the desired form.