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Exercise 6.1 · Q12

Q.Find the equation of the circle if the equations of two diameters are 2x+y=62x+y=6 and 3x+2y=43x+2y=4, when the radius of the circle is 99.

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Since both 2x+y=62x+y=6 and 3x+2y=43x+2y=4 are diameters, both pass through the centre, so the centre is their point of intersection. From 2x+y=62x+y=6, y=6−2xy=6-2x. Substituting into 3x+2y=43x+2y=4: 3x+2(6−2x)=4⇒3x+12−4x=4⇒−x=−8⇒x=83x+2(6-2x)=4 \Rightarrow 3x+12-4x=4 \Rightarrow -x=-8 \Rightarrow x=8, so y=6−16=−10y=6-16=-10. Centre …

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