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Angle of Intersection of Two Circle - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

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  • Angle of Intersection of Two Circle is considered one of the most asked concept.

  • 30 Questions around this concept.

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QuestionMEDIUM

The center of the circle \mathrm{S = 0} lie on the line \mathrm{2x-2y+9=0}  and  \mathrm{s=0} cuts orthogonally the circle \mathrm{x^{2}+y^{2}=4} . The circle \mathrm{s=0}  passes through two fixed points. Find the mid-point of their coordinates.

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If the circles \mathrm{x^2+y^2+2 x+2 k y+6=0} and \mathrm{x^2+y^2+2 k y+k=0} intersect orthogonally, the k is
 

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Two circles \mathrm{x^2+y^2+2 m x+4 y+3=0} and \mathrm{x^2+y^2+4 x-7 y+1=0} are orthogonal, then m is equal to
 

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If two circles \mathrm{(x-1)^2+(y-3)^2=r^2}  and \mathrm{x^2+y^2-8 x+2 y+8=0}  intersect in two distinct points, then
 

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The angle between the tangents drawn from the origin to the circle (x-7)^2+(y+1)^2=25 is

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If a circle passes thorough the point (1,2) and cuts the circle \mathrm{x^{2}+y^{2}=4} orthogonally, then the equation of the locus of its centre is

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The angle between the tangents drawn from the origin to the circle \mathrm{(x-7)^{2}+(y+1)^{2}=25} is

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Equation of a circle that cuts the circle \mathrm{x^2+y^2+2 g x+2 f y+c=0}, lines x = -g and y = -f orthogonally, is;

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 Tangents to the circle \mathrm{x^{2}+y^{2}=a^{2}} cut the circle \mathrm{x^{2}+y^{2}=2 a^{2}} at \mathrm{P} and \mathrm{Q}. The tangents at \mathrm{P} and \mathrm{Q} to the circle \mathrm{x^{2}+y^{2}=2 a^{2}} intersect at:

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Concepts Covered - 1

Angle of Intersection of Two Circle

Angle of Intersection of Two Circle

Angle of Intersection of Two Circle is defined as the angle between the tangents drawn to both circles at their point of intersection.

 

\\\text{Let the equation of two circle be}\\\\\begin{array}{l}{S_{1} : x^{2}+y^{2}+2 g_{1} x+2 f_{1} y+c_{1}=0} \\ {S_{2} : x^{2}+y^{2}+2 g_{2} x+2 f_{2} y+c_{2}=0}\end{array}\\\\\mathrm{C_1 \;and\; C_2\;\text{ are the centres of the given circles and} }\\\;r_1 \text{ and }r_2 \text{ are the radii of the circles.}\\\\ {\text { Thus } C_{1}=\left(-g_{1},-f_{1}\right) \text { and } C_{2}=\left(-g_{2},-f_{2}\right)} \\ {\qquad \begin{aligned}\;\;\;\;\;\;\;\; r_{1} &=\sqrt{g_{1}^{2}+f_{1}^{2}-c_{1}} \\ \text { and } &\;\;\; r_{2}=\sqrt{g_{2}^{2}+f_{2}^{2}-c_{2}} \end{aligned}}\\\\\begin{aligned} \text { Let } \quad d=C_{1} C_{2} &=\sqrt{\left(g_{1}-g_{2}\right)^{2}+\left(f_{1}-f_{2}\right)^{2}} \\ &=\sqrt{g_{1}^{2}+g_{2}^{2}+f_{1}^{2}+f_{2}^{2}-2\left(g_{1} g_{2}+f_{1} f_{2}\right)} \end{aligned}

\\ {\text{In}\;\; \Delta C_{1} P C_{2},\;\;\; \cos \alpha=\left(\frac{r_{1}^{2}+r_{2}^{2}-d^{2}}{2 r_{1} r_{2}}\right)} \\\\ {\Rightarrow \quad \cos \left(180^{\circ}-\theta\right)=\left(\frac{r_{1}^{2}+r_{2}^{2}-d^{2}}{2 r_{1} r_{2}}\right)}\\\\\mathrm{[\because \;\;\alpha+\theta+90^\circ+90^\circ=360^\circ]}\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\cos\theta=\left | \frac{r_1^2+r_2^2-d^2}{2r_1r_2} \right |}

For finding cos Ө, we are using COSINE FORMULA

 

NOTE:

If the angle of intersection two circles is 90°, then the circles are said to be orthogonal circles.

The condition for orthogonality is 2\left(g_{1} g_{2}+f_{1} f_{2}\right)=c_{1}+c_{2}

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Angle of Intersection of Two Circle

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Angle of Intersection of Two Circle Current Topic

Coordinate Axes
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Locus
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Angle of Intersection of Two Circle

Mathematics for Joint Entrance Examination JEE (Advanced) : Coordinate Geometry

Page No. : 4.30

Line : 27

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