The straight line $$x + 2y = 1$$ meets the coordinate axes at A and B. A circle is drawn through A, B and the origin. Then the sum of perpendicular distances from A and B on the tangent to the circle at the origin is:

We have the straight line $$x + 2y = 1$$. To find the points where it meets the coordinate axes, we put one coordinate equal to zero at a time.

For the $$x$$-axis we set $$y = 0$$, giving $$x + 2(0) = 1 \implies x = 1$$. Hence the point is $$A(1,\,0)$$.

For the $$y$$-axis we set $$x = 0$$, giving $$0 + 2y = 1 \implies y = \dfrac12$$. Hence the point is $$B\!\left(0,\,\dfrac12\right)$$.

The circle is required to pass through $$A(1,0)$$, $$B\!\left(0,\dfrac12\right)$$ and the origin $$O(0,0)$$. The general equation of a circle whose centre is not yet known is stated first:

$$x^{2} + y^{2} + 2gx + 2fy + c = 0.$$

Because the origin lies on the circle, substituting $$x = 0,\; y = 0$$ gives $$c = 0$$. Thus the equation reduces to

$$x^{2} + y^{2} + 2gx + 2fy = 0.$$

Next we impose the condition that $$A(1,0)$$ lies on the circle. Substituting $$x = 1,\; y = 0$$ we get

$$1^{2} + 0^{2} + 2g(1) + 2f(0) = 0 \;\;\Longrightarrow\;\; 1 + 2g = 0 \;\;\Longrightarrow\;\; g = -\dfrac12.$$

Now we impose the condition that $$B\!\left(0,\dfrac12\right)$$ lies on the circle. Substituting $$x = 0,\; y = \dfrac12$$ we get

$$0^{2} + \left(\dfrac12\right)^{2} + 2g(0) + 2f\!\left(\dfrac12\right) = 0 \;\;\Longrightarrow\;\; \dfrac14 + f = 0 \;\;\Longrightarrow\;\; f = -\dfrac14.$$

Hence the circle’s equation is

$$x^{2} + y^{2} + 2\!\left(-\dfrac12\right)x + 2\!\left(-\dfrac14\right)y = 0,$$

which simplifies to

$$x^{2} + y^{2} - x - \dfrac12\,y = 0.$$

We now determine the tangent to this circle at the origin. The standard tangent formula for a circle $$x^{2} + y^{2} + 2gx + 2fy + c = 0$$ at the point $$(x_{1},y_{1})$$ on the circle is

$$xx_{1} + yy_{1} + g(x + x_{1}) + f(y + y_{1}) + c = 0.$$

For the origin we have $$(x_{1},y_{1}) = (0,0)$$ and $$c = 0,$$ so the equation becomes simply

$$g\,x + f\,y = 0.$$

Substituting $$g = -\dfrac12$$ and $$f = -\dfrac14,$$ we obtain

$${-}\dfrac12\,x \;+\; {-}\dfrac14\,y = 0 \;\;\Longrightarrow\;\; 2x + y = 0$$

after multiplying by $$-4$$ for convenience. Thus the tangent at the origin is the straight line

$$2x + y = 0.$$

We now calculate the perpendicular distance of each of the points $$A$$ and $$B$$ from this tangent. The distance of a point $$(x_{0},y_{0})$$ from a line $$ax + by + c = 0$$ is given by the formula

$$\text{Distance} = \dfrac{|ax_{0} + by_{0} + c|}{\sqrt{a^{2} + b^{2}}}.$$

For the line $$2x + y + 0 = 0$$ we have $$a = 2,\; b = 1,\; c = 0,$$ so $$\sqrt{a^{2} + b^{2}} = \sqrt{2^{2} + 1^{2}} = \sqrt5.$$

Distance from $$A(1,0):$$}