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Then the chord of contact passes through a fixed point. Find the slope of chord of the circle having this fixed point as it’s midpoint.

Let the tangents be drawn from the point (h,k)

$$3h-4k+12=0$$

The chord of contact is $$hx+ky-4=0$$

Now$$h=\frac{4k-12}{3}$$

Then $$\frac{4k-12}{3}x+ky-4=0$$ $$4kx-12x+3ky-12=0$$ $$k(4x+3k)-12x-12=0$$ Solving the given family of lines $$x=-1$$ and $$y=\frac 43$$

The fixed point is $(-1,\frac 43)$

Slope of the line joint this point to the centre of the circle is $-\frac 43$

Then slope of chord will be $\frac 34$

But the answer given is $\frac 43$ . What’s going wrong?

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  • $\begingroup$ You are correct. In joachimsthal's notation the equation for the chord is $s_1=s_{11}$ or $-x+\frac43 y-4=(-1)^2+(\frac43)^2-4$ or $y=\frac34 x+\frac{25}{12}$ $\endgroup$ Feb 17 '20 at 13:15
  • $\begingroup$ Yup, presumably they mistyped. Good catch. $\endgroup$ Feb 17 '20 at 13:49
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It is correct to find the fixed point from the family of chord lines $k(4x+3y)-12(x+1)=0$ by requiring $4x+3y=0$ and $x+1=0$, which leads to the slope $\frac34$.

However, you may avoid the explicit equation approach, and, instead, argue that the special chord line is parallel to the given line based on the symmetry.

Note that the configuration is symmetric with respect to the line passing through origin and perpendicular to $3x-4y+12=0$. For any pair of mirror points on the given line, the midpoints of their tangent chord line are also mirror points. In the special case where the two mirror points become one on the symmetry line, so do their chord lines, which become parallel to the given line $3x-4y+12=0$ and has the same slope $\frac34$.

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  • $\begingroup$ In fact, the fixed point is the pole of the line with respect to the circle. $\endgroup$
    – amd
    Feb 17 '20 at 18:57

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