if the straight line y = x + c is tangent to the circle (x-1

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fskilnik@GMATH: GMAT Instructor; Posts: 1449; Joined: Sat Oct 09, 2010 2:16 pmFollowed by: 33 members; Thanked: 59 times

by fskilnik@GMATH » Sun Nov 25, 2018 9:30 am

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If the straight line y = x + c is tangent to the circle (x-1)^2 + (y+2)^2 = 4, what is the maximum possible value for the constant c ?

$$(A) \,\, 1 - \sqrt 2 \,\,\,\,\,\, (B) \,\, 1 + \sqrt 2 \,\,\,\,\,\, (C) \,\, 3 - 2 \sqrt 3 \,\,\,\,\,\, (D) \,\, -3 + 2 \sqrt 2 \,\,\,\,\,\, (E) \,\, -1$$

Answer: [spoiler]___(D)__[/spoiler]
Level: 650-700
Source: https://www.gmath.net

Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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Source: Beat The GMAT — Problem Solving | Sun Nov 25, 2018 9:30 am

fskilnik@GMATH: GMAT Instructor; Posts: 1449; Joined: Sat Oct 09, 2010 2:16 pmFollowed by: 33 members; Thanked: 59 times

by fskilnik@GMATH » Mon Nov 26, 2018 11:51 am

fskilnik@GMATH wrote:If the straight line y = x + c is tangent to the circle (x-1)^2 + (y+2)^2 = 4, what is the maximum possible value for the constant c ?

$$(A) \,\, 1 - \sqrt 2 \,\,\,\,\,\, (B) \,\, 1 + \sqrt 2 \,\,\,\,\,\, (C) \,\, 3 - 2 \sqrt 3 \,\,\,\,\,\, (D) \,\, -3 + 2 \sqrt 2 \,\,\,\,\,\, (E) \,\, -1$$
Source: https://www.gmath.net

\[? = {c_{\max }} = c\,\,\,\,\left( {{\rm{figure}}} \right)\]

The solution I present below is essentially geometric:

\[- 2 - c = 1 - 2\sqrt 2 \,\,\,\,\, \Rightarrow \,\,\,\,? = c = - 3 + 2\sqrt 2\]

I invite other members of this great community to present other approaches.

(I will show an algebraic solution in a couple of days, if nobody offers it before!)

Regards,
Fabio.

Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br

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regor60: Master | Next Rank: 500 Posts; Posts: 416; Joined: Thu Oct 15, 2009 11:52 am; Thanked: 27 times

by regor60 » Wed Nov 28, 2018 8:14 am

fskilnik@GMATH wrote:If the straight line y = x + c is tangent to the circle (x-1)^2 + (y+2)^2 = 4, what is the maximum possible value for the constant c ?

$$(A) \,\, 1 - \sqrt 2 \,\,\,\,\,\, (B) \,\, 1 + \sqrt 2 \,\,\,\,\,\, (C) \,\, 3 - 2 \sqrt 3 \,\,\,\,\,\, (D) \,\, -3 + 2 \sqrt 2 \,\,\,\,\,\, (E) \,\, -1$$

Answer: [spoiler]___(D)__[/spoiler]
Level: 650-700
Source: https://www.gmath.net

The center of the circle is (1, -2). The radius drawn from the center to the tangent point is perpendicular to the tangent, meaning that its slope is the negative inverse of the slope of the tangent line.

The tangent line, Y=X + C has a slope of 1, so the slope of the line from the center to the tangent is -1.

The equation of the line from the center to the tangent point can then be written as Y + 2 =-1(X - 1) or Y = -X -1.

In order to find where this line and the circle intersect, the tangent point, substitute into the circle equation

(X-1)^2 + (1 - X)^2 = 4. Expand this to X^2 -2X - 1 = 0.

Solve for X using the quadratic equation with X = 1+ or -2^1/2.

Plug this X into the circle equation to solve for Y, with Y then = 2^1/2 - 2.

So the point of tangency is (1 + or - 2^1/2, 2^1/2 - 2). This is also on the line Y =X + C. Another point on the line is where X = 0, in which case Y = C.

Using slope = (Y-Y1)/(X-X1) = 1 in this case, we have C-2^1/2+2 = 0-1+ 2^1/2 or 0-1-2^1/2.

Solve for C in both cases: C= 2(2^1/2) -3 or C = -3.

The largest C of these two choices is [spoiler]2(2^1/2) -3, D[/spoiler]

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fskilnik@GMATH: GMAT Instructor; Posts: 1449; Joined: Sat Oct 09, 2010 2:16 pmFollowed by: 33 members; Thanked: 59 times

If the straight line y = x + c is tangent to the circle (x-1

by fskilnik@GMATH » Thu Nov 29, 2018 4:12 pm

fskilnik@GMATH wrote:If the straight line y = x + c is tangent to the circle (x-1)^2 + (y+2)^2 = 4, what is the maximum possible value for the constant c ?

$$(A) \,\, 1 - \sqrt 2 \,\,\,\,\,\, (B) \,\, 1 + \sqrt 2 \,\,\,\,\,\, (C) \,\, 3 - 2 \sqrt 3 \,\,\,\,\,\, (D) \,\, -3 + 2 \sqrt 2 \,\,\,\,\,\, (E) \,\, -1$$
Source: https://www.gmath.net

\[? = {c_{\max }} = c\]

Algebraic approach:

\[\left\{ \begin{gathered}
\,\left( 1 \right)\,\,{\left( {x - 1} \right)^2} + {\left( {y + 2} \right)^2} = 4 \hfill \\
\,\left( 2 \right)\,\,y = x + c \hfill \\
\end{gathered} \right.\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( 2 \right)\,\,{\text{in}}\,\,\left( 1 \right)} \,\,\,\, \ldots \,\,\,\,\, \Rightarrow \,\,\,\,2{x^2} + 2\left( {c + 1} \right)x + {c^2} + 4c + 1 = 0\]

\[{\text{tangency}}\,\,\, \Rightarrow \,\,\,0 = \Delta = {\left[ {2\left( {c + 1} \right)} \right]^2} - 4\left( 2 \right)\left( {{c^2} + 4c + 1} \right)\,\,\, = \,\,\, \ldots \,\,\, = \,\,\, - 4\left( {{c^2} + 6c + 1} \right)\]

\[{c^2} + 6c + 1 = 0\,\,\,\,\mathop \Rightarrow \limits^{{\text{Bhaskara}}} \,\,\,\left\{ \begin{gathered}
\,{c_1} = - 3 - 2\sqrt 2 \hfill \\
\,{c_2} = - 3 + 2\sqrt 2 \hfill \\
\end{gathered} \right.\,\,\,\,\, \Rightarrow \,\,\,\,\,? = {c_2} = - 3 + 2\sqrt 2 \]

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br

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