# coordinate geometry

#### prasum

the circle x^2+y^2=a^2 cuts off intercept on the straight line lx+my=1 which subtends an angle 45 at the origin.show that a^2(l^2+m^2)=4-2*1.414

#### red_dog

Let $$\displaystyle A(x_1,y_1), \ B(x_1,y_1)$$ be the intersect points between the line and the circle. Then $$\displaystyle \widehat{AOB}=45^{\circ}$$.

The line AO has the slope $$\displaystyle m_1=\frac{y_1}{x_1}$$ and the line BO has the slope $$\displaystyle m_2=\frac{y_2}{x_2}$$

The tangent of the acute angle between the two lines is $$\displaystyle \tan\widehat{AOB}=\left|\displaystyle\frac{m_1-m_2}{1+m_1m_2}\right|$$

Replace $$\displaystyle m_1$$ and $$\displaystyle m_2$$:

$$\displaystyle 1=\tan 45^{\circ}=\left|\displaystyle\frac{x_2y_1-x_1y_2}{x_1x_2+y_1y_2}\right|$$, (1)

The coordinates of A and B are the solutions of the system formed by the equation of the line and equation of the circle.

$$\displaystyle y=\displaystyle\frac{1-lx}{m}$$.

Replace y in the equation of the circle and we have:

$$\displaystyle (m^2+l^2)x^2-2lx+1-a^2m^2=0$$

$$\displaystyle x_1, \ x_2$$ are the roots of the quadratic and we have

$$\displaystyle x_1+x_2=\displaystyle\frac{2l}{m^2+l^2}, \ x_1x_2=\displaystyle\frac{1-a^2m^2}{m^2+l^2}$$

Replace $$\displaystyle y_1,y_2$$ in (1):

$$\displaystyle 1=\displaystyle\frac{|m|\cdot|x_2-x_1|}{|(m^2+l^2)x_1x_2-l(x_1+x_2)+1|}=\frac{|m|\sqrt{(x_1+x_2)^2-4x_1x_2}}{|(m^2+l^2)x_1x_2-l(x_1+x_2)+1|}$$.

Now replace $$\displaystyle x_1+x_2$$ and $$\displaystyle x_1x_2$$.

$$\displaystyle 1=\displaystyle\frac{2\sqrt{a^2m^2+a^2l^2-1}}{2-a^2m^2-a^2l^2}\Rightarrow 2\sqrt{a^2m^2+a^2l^2-1}=1-(a^2m^2+a^2l^2-1)$$

Let $$\displaystyle \sqrt{a^2m^2+a^2l^2-1}=t$$.

Then $$\displaystyle 2t=1-t^2\Rightarrow t^2+2t-1=0$$

The positive root is $$\displaystyle t=\sqrt{2}-1$$.

$$\displaystyle \sqrt{a^2m^2+a^2l^2-1}=\sqrt{2}-1$$.

Square both members:

$$\displaystyle a^2m^2+a^2l^2-1=3-2\sqrt{2}\Rightarrow a^2(m^2+l^2)=4-2\sqrt{2}$$

Last edited by a moderator:
• prasum

#### Opalg

MHF Hall of Honor
The circle has radius $$\displaystyle a$$ and is centred at the origin. If a chord of the circle subtends an angle $$\displaystyle 45^\circ$$ at the centre of the circle then the midpoint of the chord is at a distance $$\displaystyle d = a\cos22.5^\circ$$ from the origin. But $$\displaystyle \cos^222.5^\circ = \frac{\sqrt2 +1}{2\sqrt2} = \frac1{4-2\sqrt2}$$, as you can check from the formula $$\displaystyle \frac1{\sqrt2} = \cos45^\circ = 2\cos^222.5^\circ - 1$$. It follows that $$\displaystyle d^2 = \frac{a^2}{4-2\sqrt2}$$.

The line $$\displaystyle lx+my = 1$$ intersects the circle at two points $$\displaystyle (x_1,y_1)$$, $$\displaystyle (x_2,y_2)$$ whose x-coordinates are the roots of the quadratic equation $$\displaystyle x^2 + (1 - lx)^2/m^2 = a^2$$, or $$\displaystyle (l^2+m^2)x^2 - 2lx +1-a^2m^2 = 0$$. The sum of the roots of this equation is $$\displaystyle x_1+x_2 = \frac {2l}{l^2+m^2}$$. Thus the midpoint of the chord has x-coordinate $$\displaystyle x_{\text{mid}} = \frac12(x_1+x_2) = \frac {l}{l^2+m^2}$$. Similarly, the midpoint of the chord has y-coordinate $$\displaystyle y_{\text{mid}} = \frac {m}{l^2+m^2}$$. Thus the distance of the midpoint from the origin is given by $$\displaystyle d^2 = \frac{l^2+m^2}{(l^2+m^2)^2} = \frac1{l^2+m^2}.$$

Compare the two expressions for $$\displaystyle d^2$$ to see that $$\displaystyle a^2(l^2+m^2) = 4-2\sqrt2$$.

• prasum
Similar Math Discussions Math Forum Date
Geometry
Geometry
Geometry
Geometry