mathematics-class-11-unit-12-chapter-10-circles-l-10-13-qsz2s9pdvo

### <Success>Intersection of two circles</Success>
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$S_1 : x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$, Centre $O_1(-g_1, -f_1)$
<p>$S_2 : x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0$, Centre $O_2(-g_2, -f_2)$</p>
<p>$r_1 = \sqrt{g_1^2 + f_1^2 - c_1}$, $r_2=\sqrt{g_2^2 + f_2^2 - c_2}$</p>
<p>$d_{O_1, O_2} = \sqrt{(g_2 - g_1)+(f_2 - f_1)}$</p>
<p>If $|r_1 - r_2| \leq d_{O_1, O_2} \leq r_1 + r_2$ than the two circle intersect each other.</p>
<p>If $d_{O_1, O_2} = r_1 + r_2$ than the two circles touch each other externally at exactly one point.</p>
<p>If $d_{O_1, O_2} > r_1 + r_2$ than the two circles do not intersect.</p>
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### <Success>Intersection of two circles</Success>
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<p>$r_1^2=[(c-a)+r_2 \cos \phi]^2 + [(d-b) + r_2 \sin \phi]^2$</p>
<p>$=(c-a)^2 + (d-b)^2 + r_2^2 \cos^2 \phi + r_2^2 \sin^2 \phi + 2r_2 (c-a) \cos \phi + 2r_2 (d-b) \sin \phi$</p>
<p>$r_1^2 = d_{o_1,o_2}^2 + r_2^2 + 2r_2 [(c-a) \cos \phi + (d-b) \sin \phi]$</p>
<p>$r_1^2 = d_{o_1, o_2}^2 + r_2^2 + 2r_2 d_{o_1, o_2} \left[\frac{(c-a)}{d_{o_1, o_2}} \cos \phi + \frac{(d-b)}{d_{o_1, o_2}} \sin \phi\right]$</p>
<p>$\frac{c-a}{d_{o_1, o_2}} = \cos \alpha $, $ \quad  \frac{d-b}{d_{o_1, o_2}}= \sin \alpha $</p>
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### <Success>Intersection of two circles</Success>
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$r_1^2 = d_{o_1, o_2} + r_2^2 + 2r_2 d_{o_1, 0_2} [\cos \alpha \cos \phi + \sin \alpha \sin \phi]$
<p>$ \therefore [\cos \alpha \cos \phi + \sin \alpha \sin \phi] = \cos(\phi - \alpha)$</p>
<p>$\cos(\phi - \alpha)=\frac{r_1^2 - (d_{o_1, o_2} + r_2^2)}{2r_2 d_{o_1, o_2}}$</p>
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<h3 id="successintersection-of-two-circlessuccess"><Success>Intersection of two circles</Success></h3>
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<p>$2 \pi = \phi + \beta + \frac{\pi}{2} - \alpha + \frac{\pi}{2}$</p>
<p>$\beta = \pi - \phi + \alpha$</p>








<p>$ \therefore \cos \beta = \frac{r_2^2 + d_{o_1,o_2}-r_1^2}{2r_2 d_{o_1,o_2}}$</p>
<p>$= cos ( \pi - \phi + \alpha )$</p>
<p>$ = - cos (\phi -\alpha)$</p>
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### <Success>Intersection of two circles</Success>
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<p><strong>1.)</strong> $ \quad |\frac{r_1^2 - (d_{o_1, o_2}^2 + r_2^2)}{2r_2 d_{o_1, o_2}}|<1$</p>
<p>$|r_1 - r_2|<d_{o_1,o_2}<r_1 + r_2$</p>
<p>$ d_{o_1,o_2}<r_1 + r_2$</p>
<p>$ r_1< d_{o_1, o_2} + r_2 $</p>
<p>$|r_1 - r_2|<d_{o_1,o_2}(\because r_1 > r_2)|r_2 < d_{o_1,o_2}+r_1$</p>


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