MHT CET 2023 14th May Morning Shift | Circle Question 2 | Mathematics

If the line $$x-2 y=\mathrm{m}(\mathrm{m} \in \mathrm{Z})$$ intersects the circle $$x^2+y^2=2 x+4 y$$ at two distinct points, then the number of possible values of $m$ are

MHT CET 2023 13th May Evening Shift

MCQ (Single Correct Answer)

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The abscissae of two points $$A$$ and $$B$$ are the roots of the equation $$x^2+2 a x-b^2=0$$ and their ordinates are roots of the equation $$y^2+2 p y-q^2=0$$. Then, the equation of the circle with $$A B$$ as diameter is given by

$$x^2+y^2-2 a x-2 p y+\left(b^2+q^2\right)=0$$

$$x^2+y^2-2 a x-2 p y-\left(b^2+q^2\right)=0$$

$$x^2+y^2+2 a x+2 p y+\left(b^2+q^2\right)=0$$

$$x^2+y^2+2 a x+2 p y-\left(b^2+q^2\right)=0$$

MHT CET 2023 13th May Morning Shift

MCQ (Single Correct Answer)

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The parametric equations of the curve $$x^2+y^2+a x+b y=0$$ are

$$x=\frac{\mathrm{a}}{2}+\sqrt{\frac{\mathrm{a}^2+\mathrm{b}^2}{4}} \cos \theta, y=\frac{\mathrm{b}}{2}+\sqrt{\frac{\mathrm{a}^2+\mathrm{b}^2}{4}} \sin \theta$$

$$x=\frac{\mathrm{a}}{2}-\sqrt{\frac{\mathrm{a}^2+\mathrm{b}^2}{4}} \cos \theta, y=\frac{\mathrm{b}}{2}-\sqrt{\frac{\mathrm{a}^2+\mathrm{b}^2}{4}} \sin \theta$$

$$x=-\frac{\mathrm{a}}{2}+\sqrt{\frac{\mathrm{a}^2+\mathrm{b}^2}{4}} \cos \theta, y=-\frac{\mathrm{b}}{2}+\sqrt{\frac{\mathrm{a}^2+\mathrm{b}^2}{4}} \sin \theta$$

$$x=-\frac{\mathrm{a}}{2}-\sqrt{\frac{\mathrm{a}^2+\mathrm{b}^2}{4}} \cos \theta, y=-\frac{\mathrm{b}}{2}-\sqrt{\frac{\mathrm{a}^2+\mathrm{b}^2}{4}} \sin \theta$$

MHT CET 2023 12th May Evening Shift

MCQ (Single Correct Answer)

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The circles $$x^2+y^2+2 \mathrm{a} x+\mathrm{c}=0$$ and $$x^2+y^2+2 b y+c=0$$ touch each other externally, if

$$\frac{1}{\mathrm{a}^2}-\frac{1}{\mathrm{~b}^2}=\frac{1}{\mathrm{c}}$$

$$\frac{1}{\mathrm{a}^2}+\frac{1}{\mathrm{~b}^2}=\frac{1}{\mathrm{c}}$$

$$\frac{1}{\mathrm{a}^2}+\frac{1}{\mathrm{~b}^2}=\frac{1}{\mathrm{c}^2}$$

$$\frac{1}{\mathrm{a}^2}-\frac{1}{\mathrm{~b}^2}=\frac{1}{\mathrm{c}^2}$$

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