In the -plane, a circle has equation Point is outside the circle. Exactly two distinct lines through are tangent to the circle. If the slopes of these two tangent lines are and , which choice is the value of ?

Solution available with step-by-step explanation

Super-Hard SAT Math Question 75 | Free SAT Prep | Aniko

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Question 75·200 Super-Hard SAT Math Questions·Advanced Math

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In the $xy$ -plane, a circle has equation

(x-1)^2+(y+2)^2=25.

Point $P(7,-2)$ is outside the circle. Exactly two distinct lines through $P$ are tangent to the circle. If the slopes of these two tangent lines are $m_1$ and $m_2$ , which choice is the value of $m_1m_2$ ?

For circle–line tangency problems, write the line with slope $m$ , substitute into the circle to form a quadratic in one variable, and set the discriminant to 0 to enforce exactly one intersection. If the external point is horizontally or vertically aligned with the circle’s center, the two tangents are symmetric and their slopes are opposites, making the product equal to $-m^2$ .

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Parameterize the line

Write the equation of a line with slope $m$ that passes through $P(7,-2)$ , then express $y$ (or $y+2$ ) in terms of $x$ .

Turn it into one equation in one variable

Substitute your line equation into the circle equation so you get a quadratic equation in $x$ .

Use the tangency fact

A tangent line meets the circle at exactly one point. What does that tell you about the discriminant of the quadratic in $x$ ?

Desmos Guide

Graph the circle

Enter the circle equation: (x-1)^2+(y+2)^2=25.

Graph a line through $P$ with slope slider

Enter the line y+2=m(x-7) and create a slider for m.

Find tangent slopes visually

Adjust m until the line just touches the circle (it should intersect at exactly one point). Record this slope value; then use symmetry to note the other tangent slope is the negative of it.

Compute the product and match an answer choice

Compute m*(-m) using your recorded value (or compute -(m^2)). Use the decimal you get to match the correct fraction among the answer choices.

Step-by-step Explanation

Write a tangent line through $P$ with slope $m$

Any line through $P(7,-2)$ with slope $m$ can be written as

y+2=m(x-7).

So $y=m(x-7)-2$ , and therefore $y+2=m(x-7)$ .

Substitute into the circle to get a quadratic in $x$

Substitute $y+2=m(x-7)$ into $(x-1)^2+(y+2)^2=25$ :

(x-1)^2+\big(m(x-7)\big)^2=25.

Expanding gives

(x^2-2x+1)+m^2(x^2-14x+49)-25=0,

(1+m^2)x^2+(-2-14m^2)x+(49m^2-24)=0.

Use the tangency condition (discriminant $=0$ )

A tangent line intersects the circle at exactly one point, so the quadratic in $x$ must have exactly one real solution. That happens when the discriminant is 0:

(-2-14m^2)^2-4(1+m^2)(49m^2-24)=0.

Divide by $4$ and let $t=m^2$ :

(1+7t)^2-(1+t)(49t-24)=0.

Compute each side:

$(1+7t)^2=1+14t+49t^2$
$(1+t)(49t-24)=49t^2+25t-24$

(1+14t+49t^2)-(49t^2+25t-24)=0 \Rightarrow 25-11t=0.

Find $m_1m_2$ from $m^2$

From $25-11t=0$ , we get

t=m^2=\frac{25}{11}.

Because $P(7,-2)$ lies horizontally from the circle’s center $(1,-2)$ , the two tangent lines are mirror images across the horizontal line $y=-2$ , so their slopes are $m$ and $-m$ .

Thus

m_1m_2=m(-m)=-m^2.

State the answer

Since $m^2=\frac{25}{11}$ , it follows that

m_1m_2=-\frac{25}{11}.

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Step-by-step Explanation

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