T(-2, -5) lies on the circumference of the circle with equation $(x + 3)^2 + (y + 2)^2 = 45.$ (a) Find the equation of the tangent to the circle passing through T. (b) This tangent is also a tangent to a parabola with equation $y = -2x^2 + px + 1 - p$, where $p > 3$. Determine the value of $p$.

Scottish Highers Maths Circle: T(-2, -5) lies on the circumfere

T(-2, -5) lies on the circumference of the circle with equation $(x + 3)^2 + (y + 2)^2 = 45.$ (a) Find the equation of the tangent to the circle passing through T - Scottish Highers Maths - Question 11 - 2015

Question 11

T(-2,--5)-lies-on-the-circumference-of-the-circle-with-equation--$(x-+-3)^2-+-(y-+-2)^2-=-45.$--(a)-Find-the-equation-of-the-tangent-to-the-circle-passing-through-T-Scottish Highers Maths-Question 11-2015.png

T(-2, -5) lies on the circumference of the circle with equation $(x + 3)^2 + (y + 2)^2 = 45.$ (a) Find the equation of the tangent to the circle passing through T.... show full transcript

Worked Solution & Example Answer:T(-2, -5) lies on the circumference of the circle with equation $(x + 3)^2 + (y + 2)^2 = 45.$ (a) Find the equation of the tangent to the circle passing through T - Scottish Highers Maths - Question 11 - 2015

Step 1

Find the equation of the tangent to the circle passing through T.

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Answer

To find the tangent to the circle at point T(-2, -5), we start by identifying the equation of the circle:

$(x + 3)^2 + (y + 2)^2 = 45$

The center of the circle is at C(-3, -2) and the radius can be computed as follows:

$r = ext{sqrt}(45) = 3 ext{sqrt}(5)$

Gradient of the radius: The gradient (slope) of the radius from C to T is calculated by: $m_{CT} = \frac{y_T - y_C}{x_T - x_C} = \frac{-5 - (-2)}{-2 - (-3)} = \frac{-3}{1} = -3$
Perpendicular gradient: The gradient of the tangent will be the negative reciprocal of the radius' gradient: $m_{tangent} = \frac{1}{3}$
Equation of the tangent: We can now use the point-slope form of the line equation, which is: $y - y_1 = m(x - x_1)$ Plugging in the values from T(-2, -5): $y - (-5) = \frac{1}{3}(x - (-2))$ Simplifying gives: $y + 5 = \frac{1}{3}(x + 2)$ $y = \frac{1}{3}x + \frac{2}{3} - 5$ $y = \frac{1}{3}x - \frac{13}{3}$ Thus, the equation of the tangent is: $y = \frac{1}{3}x - \frac{13}{3}$

Step 2

Determine the value of p.

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Answer

Since the tangent line must also be a tangent to the parabola given by: $y = -2x^2 + px + 1 - p$ we will set the equations equal to each other to find the values of $p$ .

Equate tangent and parabola: $\frac{1}{3}x - \frac{13}{3} = -2x^2 + px + 1 - p$ Rearranging gives: $2x^2 + (p - \frac{1}{3})x + (p - \frac{16}{3}) = 0$
Use the discriminant: To ensure there is exactly one solution (point of tangency), we set the discriminant to zero: $b^2 - 4ac = 0$ Here, using $a = 2$ , $b = p - \frac{1}{3}$ , and $c = p - \frac{16}{3}$ : $(p - \frac{1}{3})^2 - 4(2)(p - \frac{16}{3}) = 0$ Expanding gives: $p^2 - \frac{2}{3}p + \frac{1}{9} - 8p + \frac{128}{3} = 0$ This simplifies to: $p^2 - \frac{50}{3}p + \frac{385}{9} = 0$
Solve for p using the quadratic formula: $p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Thus, $p = \frac{\frac{50}{3} \pm 8}{2}$ From the calculations, we find: $p = 10$ Given the restriction $p > 3$ , this value is acceptable.

T(-2, -5) lies on the circumference of the circle with equation $(x + 3)^2 + (y + 2)^2 = 45.$ (a) Find the equation of the tangent to the circle passing through T - Scottish Highers Maths - Question 11 - 2015

Worked Solution & Example Answer:T(-2, -5) lies on the circumference of the circle with equation $(x + 3)^2 + (y + 2)^2 = 45.$ (a) Find the equation of the tangent to the circle passing through T - Scottish Highers Maths - Question 11 - 2015

Find the equation of the tangent to the circle passing through T.

Determine the value of p.

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