\( (x+5)^{2}+(y-3)^{2}=49 \) and \( y^{\prime}+x=-9 \)

We have two equations to work with: 1. The first equation is a circle: \[ (x+5)^{2}+(y-3)^{2}=49 \] This represents a circle centered at \((-5, 3)\) with a radius of \(7\) (since \(49 = 7^2\)). 2. The second equation is a linear equation: \[ y' + x = -9 \] This can be rewritten as: \[ y' = -x - 9 \] where \(y'\) represents the derivative of \(y\) with respect to \(x\). To find the points of intersection between the circle and the line, we will substitute \(y\) from the second equation into the first equation. Let's denote \(y = -x - 9\) and substitute it into the circle's equation: \[ (x+5)^{2} + ((-x - 9) - 3)^{2} = 49 \] Now, let's simplify this step by step: 1. Substitute \(y\): \[ (-x - 9 - 3) = -x - 12 \] So the equation becomes: \[ (x+5)^{2} + (-x - 12)^{2} = 49 \] 2. Expand both squares: \[ (x+5)^{2} = x^{2} + 10x + 25 \] \[ (-x - 12)^{2} = x^{2} + 24x + 144 \] 3. Combine these: \[ x^{2} + 10x + 25 + x^{2} + 24x + 144 = 49 \] \[ 2x^{2} + 34x + 169 = 49 \] 4. Move \(49\) to the left side: \[ 2x^{2} + 34x + 120 = 0 \] Now, we can solve this quadratic equation using the quadratic formula. The quadratic formula is given by: \[ x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \] where \(a = 2\), \(b = 34\), and \(c = 120\). Let's calculate the discriminant and then find the values of \(x\). Calculate the value by following steps: - step0: Calculate: \(34^{2}-4\times 2\times 120\) - step1: Multiply the terms: \(34^{2}-960\) - step2: Evaluate the power: \(1156-960\) - step3: Subtract the numbers: \(196\) Solve the equation \( 2x^2 + 34x + 120 = 0 \). Solve the quadratic equation by following steps: - step0: Solve by factoring: \(2x^{2}+34x+120=0\) - step1: Factor the expression: \(2\left(x+5\right)\left(x+12\right)=0\) - step2: Divide the terms: \(\left(x+5\right)\left(x+12\right)=0\) - step3: Separate into possible cases: \(\begin{align}&x+5=0\\&x+12=0\end{align}\) - step4: Solve the equation: \(\begin{align}&x=-5\\&x=-12\end{align}\) - step5: Rewrite: \(x_{1}=-12,x_{2}=-5\) We found the discriminant to be \(196\), which is positive, indicating that there are two real solutions for \(x\). The solutions are: \[ x_{1} = -12 \quad \text{and} \quad x_{2} = -5 \] Next, we will substitute these \(x\) values back into the equation \(y = -x - 9\) to find the corresponding \(y\) values. 1. For \(x_{1} = -12\): \[ y_{1} = -(-12) - 9 = 12 - 9 = 3 \] 2. For \(x_{2} = -5\): \[ y_{2} = -(-5) - 9 = 5 - 9 = -4 \] Thus, the points of intersection between the circle and the line are: 1. \((-12, 3)\) 2. \((-5, -4)\) In summary, the points of intersection are: - \((-12, 3)\) - \((-5, -4)\)