Our teacher told us to have a self-paced learning on pre-cal. Here's the question: Find an equation of a circle tangent to the line $x=5$ and with center $(-2,-5)$. I don't understand here is where would be the intersection at one point and find the equation.
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2$\begingroup$ Where on a circle would you draw a vertical tangent? $\endgroup$– Andrew ChinCommented Oct 20, 2020 at 7:19
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$\begingroup$ I would suggest to take some paper and make a drawing of the situation. I promise you, the problem will become much clearer at that point. $\endgroup$– Matti P.Commented Oct 20, 2020 at 7:20
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$\begingroup$ I graph it several times but I don't get the only one point $\endgroup$– HakraCommented Oct 20, 2020 at 7:25
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2 Answers
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Does this figure help you visualize the problem?
answered Oct 20, 2020 at 7:30
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welcome to MSE.
Hint: As Matti suggested draw as below:
1- Draw $y=x$, it crosses the origin O.
2- draw a perpendicular from point $C(-2, -5)$ and mark it' foot as A.
3- use the fact that tangent on a circle is perpendicular on radius of the circle. So the distance between C and A is the radius of the circle.
4- find the equation of the perpendicular:
$y-(-5) =-1[x-(-2)]$
Or:
$y=-x-7$
Now the intersection of this line with line $y=x$ give the coordinates of A as: $A(-\frac72, -\frac72)$
and the radius of circle R is:
$R=\sqrt {(-2-7/2)^2+(-5+7/2)^2}=\frac32\sqrt2$
And equation of circle is:
$(x+2)^2+(y+5)^2=R^2=\frac92$
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$\begingroup$ @Hakra, You are new to this site and you probably will seek help. But people need to be encouraged to help you. For this it is not enough to just comment and not even thank for the times people spend to answer your question. You can encourage people to answer your question in future by up-voting or accepting their solutions or helps. $\endgroup$– sirousCommented Oct 20, 2020 at 13:11