Question

Show that x+y+1=0 touches the circle x^2+y^2-3x+7y+14=0 and find its point of contact

Answer 1

Final Answer:

The equation of the straight line $x+y+1=0$ touches the equation of the circle $x^2+y^2-3x+7y+14=0$ and it happens at the point of contact (2,-3).

Given:

The equation of the straight line $x+y+1=0$ and the equation of the circle $x^2+y^2-3x+7y+14=0$ are stated.

To Find:

It is required to show that the equation of the straight line $x+y+1=0$ touches the equation of the circle $x^2+y^2-3x+7y+14=0$.

Also, the point of contact between the equation of the straight line $x+y+1=0$ and the equation of the circle $x^2+y^2-3x+7y+14=0$ is to be calculated.

Explanation:

The concepts necessary to derive the solution are the following.

• The general equation of a circle with its centre $(-g,-f)$ is $x^2+y^2+2gx+2fy+c=0$.
• The radius of the circle with its general equation$x^2+y^2+2gx+2fy+c=0$ is $r=\sqrt{g^2+f^2-c}$.
• The perpendicular distance of any point $(x_1,y_1)$ to the straight line $ax+by+c=0$ is $|\frac{ax_1+by_1+c}{\sqrt{a^2+b^2} } |$.

Step 1 of 5

As per the given data in the problem, compare the general equation $x^2+y^2+2gx+2fy+c=0$ and the given equation of the circle $x^2+y^2-3x+7y+14=0$ to derive the following.

$2g=-3\\g=-\frac{3}{2} \\\\2f=7\\f=\frac{7}{2}$

Step 2 of 5

Using the above derived information in the given problem, the following is deduced.

• The centre is

$(-g,-f)=(\frac{3}{2},-\frac{7}{2})$

• The radius is

$r\\=\sqrt{g^2+f^2-c}\\=\sqrt{(\frac{3}{2})^2+(-\frac{7}{2})^2-14}\\=\sqrt{\frac{9}{4}+\frac{49}{4}-14}\\=\sqrt{\frac{9+49-4\times 14}{4}}\\\\=\frac{1}{\sqrt{2}}$.

Step 3 of 5

In continuation with the above calculations, the perpendicular distance of the centre $(\frac{3}{2},-\frac{7}{2})$ from the straight line $x+y+1=0$ is

$|\frac{\frac{3}{2}-\frac{7}{2}+1}{\sqrt{1^2+1^2} } |\\\\=|\frac{\frac{3-7+2}{2}}{\sqrt{2}} |\\=|\frac{-\frac{2}{2}}{\sqrt{2}} |\\\\=\frac{1}{\sqrt{2}}$

Since the distance of the centre $(\frac{3}{2},-\frac{7}{2})$ from the straight line $x+y+1=0$ is equal to the radius, the equation of the straight line $x+y+1=0$ touches the equation of the circle $x^2+y^2-3x+7y+14=0$.

Step 4 of 5

Substitute the value of y from the equation of the straight line $x+y+1=0$  in the equation of the circle $x^2+y^2-3x+7y+14=0$.

$x^2+y^2-3x+7y+14=0\\x^2+(-1-x)^2-3x+7(-1-x)+14=0\\x^2+(x+1)^2-3x-7-7x+14=0\\2x^2+2x+1-10x+7=0\\2x^2-8x+8=0\\x^2-4x+4=0\\(x-2)^2=0\\x=2,2$

Step 5 of 5

Substitute this value of x in the equation of the straight line $x+y+1=0$ to get the value of y.

$y\\=-1-2\\=-3$

From the above calculations, the point of contact is (2,-3).

Therefore, the required correct answer is the point of contact between  $x+y+1=0$ and $x^2+y^2-3x+7y+14=0$ is (2,-3).

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