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Find the equation of the circle of radius $5$ units which lies within the circle $x^2 + y^2 + 14x + 10y - 26 = 0$ and which touches the given circle at the point $(-1,3)$.

Please help me.

Thanks in Advance.

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  • $\begingroup$ Can you write the equation of the outer circle in standard form, $(x-x_c)^2+(y-y_c)^2=r^2$ $\endgroup$
    – Ross Millikan
    Jan 8, 2017 at 16:01

2 Answers 2

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Hints:

$$x^2+y^2+14x+10y-26=0\iff (x+7)^2+(y+5)^2=100$$

The line through $\;(-1,3)\;$ and the big circle's center passes through the little circle's center (why?), so it is on the line

$$y-2=\frac86(x+1)$$

Finally, you need the point on the above line at a distance of $\;5\;$ from $\;(-7,-5)\;$ and between $\;(-7,-5)\;$ and $\;(-1,3)\;$ .

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  • $\begingroup$ Will you please elaborate a little: Big Circles center passes through little circle's center. ?? $\endgroup$
    – Vidyanshu Mishra
    Jan 8, 2017 at 16:10
  • $\begingroup$ @THELONEWOLF. This is a theorem in geometry: when we have two tangential circles, either externally or internally, the line determined by the circles' centers passes through the tangency point...and I didn't write the big circle's center passes (??) throught the little one's center. Read the complete sentence carefully. $\endgroup$
    – DonAntonio
    Jan 8, 2017 at 16:18
  • $\begingroup$ ooh... SOrry... @Don.. Actually it become quite clear on a careful reading. Thanks and apologies. $\endgroup$
    – Vidyanshu Mishra
    Jan 8, 2017 at 16:22
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Hint:

note that the center of the given circle is $C=(-7,-5)$ and the radius is $r=10$, so the center of the internal circle is the midpoint from $(-7,-5)$ and $(-1,3)$.

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