Prove that the lines joining the origin to the points of intersection of circle x-h2-y-k2-k2 with the line xh-yk-2 will be perpendicular if h- k lies on the circle x2-y2-r2-

Making the equation of circle homogeneous, x^2+y^2 2(hx+ky)·12(xh+yk)=(h^2+k^2 r^2)·14(xh+yk)^2 A+B=0⇒(h^2+k^2 r^24)(1h^2+1k^2)=0 ∴ h^2+k^2 r^2=0 ...

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Standard X

Mathematics

Two Circles Touching Internally and Externally

Prove that th...

Question

Prove that the lines joining the origin to the points of intersection of circle $(x - h)^{2} + (y - k)^{2} = k^{2}$ with the line $\frac{x}{h} + \frac{y}{k} = 2$ will be perpendicular if (h, k) lies on the circle $x^{2} + y^{2} = r^{2}$ .

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Solution

Making the equation of circle homogeneous,

$x^{2} + y^{2} - 2 (h x + k y) \cdot \frac{1}{2} (\frac{x}{h} + \frac{y}{k}) = (h^{2} + k^{2} - r^{2}) \cdot \frac{1}{4} {(\frac{x}{h} + \frac{y}{k})}^{2}$

$A + B = 0 \Rightarrow (\frac{h^{2} + k^{2} - r^{2}}{4}) (\frac{1}{h^{2}} + \frac{1}{k^{2}}) = 0$

$∴ h^{2} + k^{2} - r^{2} = 0$ , as $\frac{1}{h^{2}} + \frac{1}{k^{2}} \neq 0$

$∴$ Locus of (h, k) is $x^{2} + y^{2} = r^{2}$

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Prove that the lines joining the origin to the points of intersection of circle (x−h)2+(y−k)2=k2 with the line xh+yk=2 will be perpendicular if (h, k) lies on the circle x2+y2=r2.

Making the equation of circle homogeneous,x2+y2−2(hx+ky)⋅12(xh+yk)=(h2+k2−r2)⋅14(xh+yk)2A+B=0⇒(h2+k2−r24)(1h2+1k2)=0∴h2+k2−r2=0, as 1h2+1k2≠0∴ Locus of (h, k) is x2+y2=r2

Prove that the lines joining the origin to the points of intersection of circle $(x - h)^{2} + (y - k)^{2} = k^{2}$ with the line $\frac{x}{h} + \frac{y}{k} = 2$ will be perpendicular if (h, k) lies on the circle $x^{2} + y^{2} = r^{2}$ .