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The circle c has equation $x^2 + y^2 - 2x + 8y + k = 0$, where $k \in \mathbb{R}$ - Leaving Cert Mathematics - Question 3 - 2022

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The circle c has equation $x^2 + y^2 - 2x + 8y + k = 0$, where $k \in \mathbb{R}$. The radius of c is $5\sqrt{3}$. Find the value of $k$. The circle $(x - 5)^2 + (... show full transcript

Worked Solution & Example Answer:The circle c has equation $x^2 + y^2 - 2x + 8y + k = 0$, where $k \in \mathbb{R}$ - Leaving Cert Mathematics - Question 3 - 2022

Step 1

Find the value of k

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Answer

To find the value of kk, we start from the standard form of a circle: R=r2R = \sqrt{r^2} where RR is the radius. Given that the radius is 535\sqrt{3}, we have:

R=53R = 5\sqrt{3}

The radius can also be expressed using the formula for the equation of the circle:

R=(1−22)2+(4−82)2R = \sqrt{\left(1 - \frac{2}{2}\right)^2 + \left(4 - \frac{8}{2}\right)^2}.

First, we simplify the equation:

12+42−k=R21^2 + 4^2 - k = R^2 This simplifies to: 1+16−k=751 + 16 - k = 75 Thus: 17−k=75\n⇒k=17−75\n⇒k=−5817 - k = 75\n \Rightarrow k = 17 - 75\n \Rightarrow k = -58

Step 2

Find the slope of this tangent

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Answer

To find the slope of the tangent to the circle at the point (9,−4)(9, -4), we use implicit differentiation:

  1. Differentiate both sides of the circle equation: (x−5)2+(y−2)2=20(x - 5)^2 + (y - 2)^2 = 20

  2. Applying the chain rule gives: 2(x−5)+2(y−2)dydx=02(x - 5) + 2(y - 2) \frac{dy}{dx} = 0

  3. Isolate dydx\frac{dy}{dx}: dydx=−x−5y−2\frac{dy}{dx} = -\frac{x - 5}{y - 2}

  4. Substitute the point (9,−4)(9, -4) into this equation: dydx=−9−5−4−2=−4−6=23\frac{dy}{dx} = -\frac{9 - 5}{-4 - 2} = -\frac{4}{-6} = \frac{2}{3}

Thus, the slope of the tangent line at the point (9,−4)(9, -4) is 23\frac{2}{3}.

Step 3

Find the equation of each of these circles

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Answer

To find the equation of the circles, we start with its center (r,−r)(r, -r):

  1. Using the point (1,−8)(1, -8), we can find the relationship: (1−r)2+(−8+r)2=r2(1 - r)^2 + (-8 + r)^2 = r^2

  2. Expanding the equation gives: (1−r)2=1−2r+r2(1 - r)^2 = 1 - 2r + r^2 (−8+r)2=64−16r+r2(-8 + r)^2 = 64 - 16r + r^2

  3. Summing these gives: 1−2r+r2+64−16r+r2=r21 - 2r + r^2 + 64 - 16r + r^2 = r^2 65−18r+r2=r265 - 18r + r^2 = r^2 65−18r=0\n⇒r=651865 - 18r = 0\n \Rightarrow r = \frac{65}{18}

Therefore, the equations of the circles can be derived as:

(x - r)^2 + (y + r)^2 &= r^2 \ (x + r)^2 + (y - r)^2 &= r^2 \end{align*}$$ The final equations will incorporate the calculated values of $r$.

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