Statement (A) : For all values of theta,the l | vedclass

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(A) The general equation of a tangent to the circle $x^2 + y^2 = a^2$ at any point $(a \cos 	heta, a \sin 	heta)$ is given by $x \cos 	heta + y \si

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Both $(A)$ and $(R)$ are true and $(R)$ is the correct explanation of $(A)$.

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(A) The general equation of a tangent to the circle $x^2 + y^2 = a^2$ at any point $(a \cos 	heta, a \sin 	heta)$ is given by $x \cos 	heta + y \sin 	heta = a$.For the circle $(x - 3)^2 + (y - 3)^2 = 1$,we can perform a translation of axes where $X = x - 3$ and $Y = y - 3$.The equation becomes $X^2 + Y^2 = 1^2$,which is of the form $X^2 + Y^2 = a^2$ with $a = 1$.The tangent to this circle is $X \cos 	heta + Y \sin 	heta = 1$.Substituting back,we get $(x - 3) \cos 	heta + (y - 3) \sin 	heta = 1$.Thus,both $(A)$ and $(R)$ are true,and $(R)$ provides the correct explanation for $(A)$.

Statement $(A) :$ For all values of $\theta$,the line $(x - 3) \cos \theta + (y - 3) \sin \theta = 1$ is tangent to the circle $(x - 3)^2 + (y - 3)^2 = 1$.
Reason $(R) :$ For all values of $\theta$,the line $x \cos \theta + y \sin \theta = a$ is tangent to the circle $x^2 + y^2 = a^2$.

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Statement $(A) :$ For all values of $\theta$,the line $(x - 3) \cos \theta + (y - 3) \sin \theta = 1$ is tangent to the circle $(x - 3)^2 + (y - 3)^2 = 1$.Reason $(R) :$ For all values of $\theta$,the line $x \cos \theta + y \sin \theta = a$ is tangent to the circle $x^2 + y^2 = a^2$.