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  • The circles  <img alt="\mathrm{x^2+y^2-4 x-6 y+9=0 \text { and } x^2+y^2+2 x+2 y-7=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bx%5E2&plus;y%5E2-4%20x-6%20y&plu

The circles  \mathrm{x^2+y^2-4 x-6 y+9=0 \text { and } x^2+y^2+2 x+2 y-7=0}  touch. Find the coordinates of the point of contact.

 

Option: 1

\mathrm{\left(\frac{4}{5}, \frac{7}{5}\right)}


Option: 2

\mathrm{\left(\frac{4}{3}, \frac{7}{5}\right) }


Option: 3

\mathrm{\left(-\frac{4}{5}, \frac{7}{3}\right)}


Option: 4

none of these


Answers (1)

best_answer

 The centres of the given circles are A(2, 3) and B(–1, –1). The radii of the circles are 


\mathrm{r_1=\sqrt{4+9-9}=2 \text { and } r_2=\sqrt{1+7+7}=3 \text {. Furthermore } A B=\sqrt{(2+1)^2+(3+1)^2}=5 \text {. }}

since \mathrm{\mathrm{r}_1+\mathrm{r}_2=2+3=5=\mathrm{AB}} the circles touch externally at a point \mathrm{\mathrm{P}(\alpha, \beta)} Furthermore AP : PB = 2 : 3 and

so \mathrm{\alpha=\frac{2(-1)+3(2)}{2+3}=\frac{4}{5}, \beta=\frac{2(-1)+3(3)}{2+3}=\frac{7}{5}}

 i.e. P is the point (4/5, 7/5)

 

Posted by

Ritika Harsh

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