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# Location of the circle in relation to a circle Assignment Help

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Circle - Location of the circle in relation to a circle
**Location of the circle in relation to a circle:**

Let S_{1} ≡ x^{2} + y^{2} + 2g_{1}x + 2*f*_{1}y+ c_{1} = 0 and S_{2} º x^{2} + y^{2} + 2g_{2}x + 2*f*_{2}y+ c_{1} = 0 be two circles. Let D be discriminant for the quadratic equation in x (or y) which is obtained by eliminating y (or x) from the 2 equations of the circle. Then

(i) they are 2 intersecting circles if D > 0

(ii) they are nonintersecting (having no common points) if D < 0

(iii) they touch each other if D = 0

(iv) If D < 0, that is, the circles are nonintersecting then

(a) S_{1} = 0 is outside S_{2} = 0 if S_{2} (-g_{1}, -*f*_{1}) > 0 or S_{1} (-g_{2}, -*f*_{2}) > 0; equivalently, AB > r_{1} + r_{2} here A, B are centres and r_{1}, r_{2} are radii respectively.

(b) S_{1} = 0 is inside S_{2} = 0 if S_{2} (-g_{1}, -*f*_{1}) < 0; equivalently, AB < |r_{2} - r_{1}|

(v) If D = 0, then the circles touches each other

(c) externally if AB = r_{1} + r_{2}

(d) internally if AB = | r_{1} - r_{2}|

**Chord of contact**

From the point P(x_{1}, y_{1}) outside the circle 2 tangents PA and PB can be drawn to circle. The chord AB joining the points of contact A and B of the tangents from P is called as chord of contact of P(x_{1}, y_{1}) with respect to circle. Its equation can be given by T = 0.

*Illustration: If chord of contact of the tangents drawn to x*^{2}+y^{2}=a^{2} from any point on x^{2}+y^{2}=b^{2}, touches the circle x^{2}+y^{2}=c^{2}, then show that a^{2}=bc

*Solution: *Assume that P(x_{1}, y_{1}) be any point on x^{2}+y^{2}=b^{2} that is x_{1}^{2}+y_{1}^{2}=b^{2}. Equation of the corresponding chord of contact is xx_{1}+yy_{1}-a^{2}=0. It

*Solution:*

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**Location of the circle in relation to a circle:**

_{1}≡ x

^{2}+ y

^{2}+ 2g

_{1}x + 2

*f*

_{1}y+ c

_{1}= 0 and S

_{2}º x

^{2}+ y

^{2}+ 2g

_{2}x + 2

*f*

_{2}y+ c

_{1}= 0 be two circles. Let D be discriminant for the quadratic equation in x (or y) which is obtained by eliminating y (or x) from the 2 equations of the circle. Then

_{1}= 0 is outside S

_{2}= 0 if S

_{2}(-g

_{1}, -

*f*

_{1}) > 0 or S

_{1}(-g

_{2}, -

*f*

_{2}) > 0; equivalently, AB > r

_{1}+ r

_{2}here A, B are centres and r

_{1}, r

_{2}are radii respectively.

_{1}= 0 is inside S

_{2}= 0 if S

_{2}(-g

_{1}, -

*f*

_{1}) < 0; equivalently, AB < |r

_{2}- r

_{1}|

_{1}+ r

_{2}

_{1}- r

_{2}|

**Chord of contact**

From the point P(x_{1}, y_{1}) outside the circle 2 tangents PA and PB can be drawn to circle. The chord AB joining the points of contact A and B of the tangents from P is called as chord of contact of P(x_{1}, y_{1}) with respect to circle. Its equation can be given by T = 0.

*Illustration: If chord of contact of the tangents drawn to x*^{2}+y^{2}=a^{2}from any point on x^{2}+y^{2}=b^{2}, touches the circle x^{2}+y^{2}=c^{2}, then show that a^{2}=bc**Assume that P(x**

*Solution:*_{1}, y

_{1}) be any point on x

^{2}+y

^{2}=b

^{2}that is x

_{1}

^{2}+y

_{1}

^{2}=b

^{2}. Equation of the corresponding chord of contact is xx

_{1}+yy

_{1}-a

^{2}=0. It

*Solution:***Email based Location of the circle in relation to a circle Assignment Help - Homework Help**

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