x^2+y^2+2x+4y-20=0 and x^2+y^2+6x-8y+10=0 are | vedclass

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(A) The equations of the given circles are $x^2+y^2+2x+4y-20=0$ and $x^2+y^2+6x-8y+10=0$. Comparing with $x^2+y^2+2g_1x+2f_1y+c_1=0$ and $x^2+y^2+2g_2

Vedclass · Accepted Answer

They intersect orthogonally and will have two common tangents. The length of their common chord is $\frac{5\sqrt{3}}{\sqrt{2}}$

Vedclass · Answer

(A) The equations of the given circles are $x^2+y^2+2x+4y-20=0$ and $x^2+y^2+6x-8y+10=0$. Comparing with $x^2+y^2+2g_1x+2f_1y+c_1=0$ and $x^2+y^2+2g_2x+2f_2y+c_2=0$,we get $g_1=1, f_1=2, c_1=-20$ and $g_2=3, f_2=-4, c_2=10$. Condition for orthogonality is $2g_1g_2+2f_1f_2 = c_1+c_2$. $2(1)(3)+2(2)(-4) = 6-16 = -10$ and $c_1+c_2 = -20+10 = -10$. Since $2g_1g_2+2f_1f_2 = c_1+c_2$,the circles intersect orthogonally and have two common tangents. The equation of the common chord is $(x^2+y^2+2x+4y-20) - (x^2+y^2+6x-8y+10) = 0$,which simplifies to $-4x+12y-30=0$ or $2x-6y+15=0$. The center of the first circle is $C_1(-1, -2)$ and its radius is $r_1 = \sqrt{1^2+2^2-(-20)} = \sqrt{25} = 5$. The length of the common chord is $2\sqrt{r_1^2-d^2}$,where $d$ is the distance from $C_1$ to the chord $2x-6y+15=0$. $d = \frac{|2(-1)-6(-2)+15|}{\sqrt{2^2+(-6)^2}} = \frac{|-2+12+15|}{\sqrt{40}} = \frac{25}{\sqrt{40}} = \frac{25}{2\sqrt{10}}$. Length $= 2\sqrt{25 - \frac{625}{40}} = 2\sqrt{\frac{1000-625}{40}} = 2\sqrt{\frac{375}{40}} = 2\sqrt{\frac{75}{8}} = 2 \cdot \frac{5\sqrt{3}}{2\sqrt{2}} = \frac{5\sqrt{3}}{\sqrt{2}}$.

$x^2+y^2+2x+4y-20=0$ and $x^2+y^2+6x-8y+10=0$ are the given circles. Which one of the following is correct?

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