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In the given figure, $R$ is the center of the circle. The circle touches $X$ axis at $S(7,0)$ and intercepts the $y$ axis with $PQ=48$ units. From this information, find the equation of the circle. enter image description here

My Attempt:
Let $R(h,k)$ be the center of the circle. Then $h=7$. Let the equation of circle be $$x^2+y^2+2gx+2fy+c=0$$ The circle passes through $(7,0)$, so $$(7)^2+(0)^2+2g(7)+2f(0)+ c=0$$ $$49+0+14g +0+c=0$$ How do I proceed from here?

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3 Answers 3

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Let $M$ be the midpoint of $PQ$, then we have the right triangle

  P
  |\
24| \
  |  \
  M---R
    7

$RP$ is the circle's radius and is equal to $\sqrt{24^2+7^2}=25$. Since $RS=RP$, $R=(7,25)$ and the equation of the circle is $(x-7)^2+(y-25)^2=625$.

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Let $T$ be the midpoint of the line segment $PQ$.

Now consider the right triangle $PTR$ :

$$\text{(the radius)}=PR=\sqrt{PT^2+TR^2}=\sqrt{24^2+7^2}=25$$

Also, note that the $y$-coordinate of $R$ is the same as the radius.

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As $R (h, k) $ is the centre of the circle, we can write the equation as $$(x-7)^2 +(y-k)^2 =r^2 \tag {1}$$ where $r $ is the radius of the circle and we know that the abscissa of the centre is $7$.

Also as the circle passes through $P $ and $Q$ which are of the form $(0, \alpha) $ where $\alpha $ is some number, these two points must satisfy $(1)$. Thus, we have $(1) $ at $P $ and $Q $ as: $$49 + (\alpha-k)^2 =r^2$$ $$\Rightarrow \alpha = k \pm \sqrt {r^2-49} $$ Let the two values of this $y $ be $\alpha_1$ and $\alpha_2$. Given $$\alpha_1-\alpha_2 =48 \implies r = 25 \tag {2}$$

Also $S (7,0) $ satisfies $(1) $ giving us $k =25$. Fill up the necessary details in $(1) $ and the result follows. Hope it helps.

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  • $\begingroup$ @@Rohan, why is it $(y-\alpha) ^2$ but not $(\alpha -k)^2$?? $\endgroup$
    – pi-π
    Feb 2, 2017 at 15:22
  • $\begingroup$ @NeWtoN Thanks for correcting the typo. $\endgroup$
    – user371838
    Feb 2, 2017 at 15:34

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