MHT CET 2023 13th May Evening Shift | Circle Question 3 | Mathematics

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

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

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

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

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

MHT CET 2023 12th May Morning Shift

MCQ (Single Correct Answer)

-0

If $$\lambda$$ is the perpendicular distance of a point $$\mathrm{P}$$ on the circle $$x^2+y^2+2 x+2 y-3=0$$, from the line $$2 x+y+13=0$$, then maximum possible value of $$\lambda$$ is

$$2 \sqrt{5}$$

$$3 \sqrt{5}$$

$$4 \sqrt{5}$$

$$\sqrt{5}$$

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