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Square MATH has sides of length 4, and N is the midpoint of TH. A circle with radius 2 and center N intersects a circle with radius 4 and center M at points P and H. What is the distance from P to MH.

 Feb 17, 2020
 #1
avatar+531 
+2

Square MATH has sides of length 4, and N is the midpoint of TH. A circle with radius 2 and center N intersects a circle with radius 4 and center M at points P and H. What is the distance from P to MH.

 

MH = 4

HN = 2

Angles HMN and PMN are equal.

tan(∠ HMN) = HN / MH = 26.565°

Angle HMP = (∠ HMN) * 2 = 53.13°

LP => the distance from P to MH

LP = sin(∠ HMN) * MP          LP = 3.2  indecision

 

 Feb 17, 2020
edited by Dragan  Feb 17, 2020
edited by Dragan  Feb 17, 2020
edited by Dragan  Feb 18, 2020
 #2
avatar+11731 
+2

Square MATH has sides of length 4, and N is the midpoint of TH. A circle with radius 2 and center N intersects a circle with radius 4 and center M at points P and H. What is the distance from P to MH.

laugh

 Feb 17, 2020
 #3
avatar+24430 
+2

Square MATH has sides of length 4, and N is the midpoint of TH.
A circle with radius 2 and center N intersects a circle with radius 4 and center M at points P and H.
What is the distance from P to MH.

 

\(\begin{array}{|rcll|} \hline \mathbf{\text{Circle with }r=4 \\ \text{ and center } M(0,4):} \\ \hline x^2+(y-4)^2 &=& 4^2 \\ \hline \end{array} \begin{array}{|rcll|} \hline \mathbf{\text{Circle with }r=2 \\ \text{ and center } N(2,0):} \\ \hline (x-2)^2+y^2 &=& 2^2 \\ \hline \end{array} \)

 

\(\begin{array}{|lcll|} \hline \mathbf{\text{Intersects at }P(x_P,\ y_P):} \\ \begin{array}{|rcll|} \hline x_P^2+(y_P-4)^2 &=& 16 \\ x_P^2+y_P^2-8yP + 16 &=& 16 \\ x_P^2+y_P^2-8yP &=& 0 \\ \mathbf{x_P^2+y_P^2} &=& \mathbf{8y_P} \\ \hline \end{array} \begin{array}{|rcll|} \hline (x_P-2)^2+y_P^2 &=& 4 \\ x_P^2-4x_P + 4 +y_P^2 &=& 4 \\ x_P^2 +y_P^2-4x_P &=& 0 \\ \mathbf{x_P^2 +y_P^2}&=& \mathbf{4x_P} \\ y_P^2 &=& 4x_P -x_P^2 \\ \mathbf{ y_P } &=& \mathbf{\sqrt{ 4x_P -x_P^2}} \\ \hline \end{array} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline \mathbf{x_P^2+y_P^2} &=& \mathbf{8y_P} \quad | \quad \mathbf{x_P^2 +y_P^2=4x_P},\ \mathbf{ y_P =\sqrt{ 4x_P -x_P^2}} \\ 4x_P &=& 8\sqrt{ 4x_P -x_P^2} \quad | \quad :4 \\ x_P &=& 2\sqrt{ 4x_P -x_P^2} \quad | \quad \text{square both sides} \\ x_P^2 &=& 4(4x_P -x_P^2) \\ x_P^2 &=& 16x_P -4x_P^2 \\ 5x_P^2 &=& 16x_P \\ 5x_P^2-16x_P &=& 0 \\ x_P(5x_P-16) &=& 0 \quad | \quad x_P \neq 0~! \\\\ 5x_P-16 &=& 0 \\ 5x_P &=& 16 \\ x_P &=& \dfrac{16}{5} \\ \mathbf{x_P} &=& \mathbf{3.2} \\ \hline \end{array}\)

 

The distance from P to MH is \(\mathbf{3.2}\)

 

laugh

 Feb 17, 2020

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