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Byju's Answer

Standard XII

Mathematics

Parametric Form of Normal : Ellipse

Circles C 1 a...

Question

Circles $C_{1}$ and $C_{2}$ , of radii $r$ and $R$ respectively, touch each other as shown in the figure. The line $A$ , which is parallel to the line joining the centres of $C_{1}$ and $C_{2}$ , is tangent to $C_{1}$ at $P$ and intersects $C_{2}$ at $A, B$ . If $R^{2} = 2 r^{2}$ , then $∠ A O B$ equals

22 {\frac{1}{2}}^{\circ}

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45^{\circ}

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60^{\circ}

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67 {\frac{1}{2}}^{\circ}

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Solution

The correct option is B $45^{\circ}$

Equation of $A B \Rightarrow y = r$
Equation of circle $C_{2} \Rightarrow (x - R)^{2} + y^{2} = R^{2}$
So, $point A (R - \sqrt{R^{2} - r^{2}}, r) Using R^{2} = 2 r^{2}$
$point B (R + \sqrt{R^{2} - r^{2}}, r)$
$\Rightarrow A (R - r, r), B (R + r, r)$
Slope of $O A = \frac{r}{R - r} = m_{1}$
Slope of $O B = \frac{r}{R + r} = m_{2}$
$tan θ = \frac{m_{1} - m_{2}}{1 + m_{1} . m_{2}} = 1 ∴ θ = 45^{\circ}$

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Circles C1 and C2 , of radii r and R respectively, touch each other as shown in the figure. The line A, which is parallel to the line joining the centres of C1 and C2 , is tangent to C1 at P and intersects C2 at A,B. If R2=2r2, then ∠AOB equals

The correct option is B 45∘ Equation of AB⇒y=r Equation of circle C2⇒(x−R)2+y2=R2 So,point A(R−√R2−r2,r) Using R2=2r2 point B(R+√R2−r2,r) ⇒A(R−r,r), B(R+r,r) Slope of OA=rR−r=m1 Slope of OB=rR+r=m2 tanθ=m1−m21+m1.m2=1∴θ=45∘