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Question

If the chord of contact of tangents from a point P to a given circle passes through Q, then the circle on PQ as diameter.



A
cuts the given circle orthogonally
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B
touches the given circle externally
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C
touches the given circle internally
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D
none of these
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Solution

The correct option is A cuts the given circle orthogonally
Let S:$$x^2+y^2+2gx+2fy+c=0$$ be the circle'

Chord of contact from P($$x_1,y_1) $$is given by,

$$xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0$$

Since $$Q(x_2,y_2) $$ lies on the above line

$$x_1x_2+y_1y_2+g(x_1+x_2)+f(y_1+y_2)+c=0~~~~~~-(1)$$

Centre of $$S_1=(-g,-f), r_1=\sqrt{g^2+f^2-c}$$

Midpoint of PQ=$$C-2(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2})$$


$$r_2=\dfrac{\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}}{2}$$


Circle with centre $$c_2$$ and radius $$r_2$$ is given by

$$S_2:(x-(\dfrac{x_1+x_2}{2}))^2+(y-9\dfrac{y_1+y_2}{2}))^2=r_2^2$$

$$C_1C_2=\sqrt{(\dfrac{x_1+x_2}{2}+g)^2+(\dfrac{y_1+y_2}{2}+f)^2}$$


Simplifying above equation usong (1)

$$C_1C_2^2=\dfrac{(x_1-x_2)^2}{4}+\dfrac{(y_1-y_2)^2}{4}+g^2+f^2-c$$

$$r_1^2+r_2^2=(g^2+f^2-c)+\dfrac{(x_1-x_3)^2+(y_1-y_2)^2}{4}$$

$$\implies C_1C_2^2=r_1^2+r_2^2$$

$$S_1$$ abd $$S_2$$ cut orthogonally.


Mathematics

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