# Find the equation of the circle given the tangent line, point on the circle, and the radius [closed]

The problem was this: radius is $$2$$, tangent to the $$x$$ axis, passes through $$(1, -1)$$. I don't know how to solve this, and my math teacher didnt teach this yet.

## closed as off-topic by Namaste, Brahadeesh, Paul Frost, user296602, José Carlos SantosNov 18 '18 at 23:29

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• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Namaste, Brahadeesh, Paul Frost, Community, José Carlos Santos
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• "my math teacher didnt teach this yet" A teacher shouldn't have to teach the exact method of solving a specific problem in order to expect his / her students to solve it. You have likely been taught all the tools you need: You know what the equation of a circle looks like, and you've been taught how to solve equations. In this problem you've been given information you can enter directly into that equation in order to solve and figure out the parts you haven't been told. Try it out, see how far you get, then let us know where you're stuck. – Arthur Nov 18 '18 at 10:25
• The circle also passes through $(x,0)$. – Yadati Kiran Nov 18 '18 at 10:49
• $(x-h)^2 + (y-k)^2=2^2$ How to find $(h,k)?$ – Narasimham Nov 18 '18 at 18:17
• Hint: the center has coordinates $(a,2)$ or $(a,-2)$. – egreg Nov 18 '18 at 23:30

Given the conditions you gave, there is only two circle that works: their center is $$\left(\begin{array}{cc} 1-\sqrt{3}\\ -2 \end{array}\right)$$ and $$\left(\begin{array}{cc} 1+\sqrt{3}\\ -2 \end{array}\right)$$, their radius is $$2$$.

So their respective equation are:

$$(x - 1+\sqrt{3})^2 + (y+2)^2 = 4$$ and $$(x-1-\sqrt{3})^2 + (y+2)^2 = 4$$.

Since the circle is tangent to the x-axis and passes through $$(1,-1)$$ with radius $$2$$ we know the center is below the x-axis and $$2$$ units away from it.

Thus the center is a point $$C(x,-2)$$ which is $$2$$ unit apart from $$(1,-1)$$

That gives us $$(x-1)^2 + (-1)^2 =4$$ which implies $$x=1\pm \sqrt 3$$

There are two circles with equations $$(x-1\pm \sqrt 3)^2 +(y+2)^2=4$$