You visited us 1 times! Enjoying our articles? Unlock Full Access!

Chord of Contact

Let x-2y+7=0 ...

Question

Let $x - 2 y + 7 = 0$ be a chord of the circle $x^{2} + y^{2} - 2 x - 10 y + 1 = 0$ . If the midpoint of the chord is $P (α, β)$ , then the value of $5 | α - β |$ is

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

14.0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

14.00

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

Given chord is $x - 2 y + 7 = 0$
$P = (α, β)$
Now, $α - 2 β + 7 = 0 \dots (1)$
$C P$ is perpendicular to chord, so
$\frac{β - 5}{α - 1} = - 2$
Using equation $(1)$ , we get
$β - 5 = - 2 (2 β - 8) \Rightarrow β = \frac{21}{5} \Rightarrow α = \frac{7}{5} ∴ 5 | α - β | = 14$

Alternate method :
Given circle is $x^{2} + y^{2} - 2 x - 10 y + 1 = 0$
Centre and radius is $C = (1, 5), r = 5$
$P$ is foot of perpendicular drawn from $(1, 5)$ to $x - 2 y + 7 = 0$ , so
$\frac{α - 1}{1} = \frac{β - 5}{- 2} = - (\frac{1 - 10 + 7}{1^{2} + 2^{2}}) \Rightarrow \frac{α - 1}{1} = \frac{β - 5}{- 2} = \frac{2}{5} \Rightarrow α = \frac{7}{5}, β = \frac{21}{5} ∴ 5 | α - β | = 14$

Suggest Corrections

Similar questions

x - 2 y + 7 = 0

x^{2} + y^{2} - 2 x - 10 y + 1 = 0

P (α, β)

5 | α - β |

x - 2 y + 7 = 0

x^{2} + y^{2} - 2 x - 10 y + 1 = 0

P (α, β)

5 | α - β |

x - 2 y + 7 = 0

x^{2} + y^{2} - 2 x - 10 y + 1 = 0

The length and the midpoint of the chord 4x-3y+5=0 w.r.t circle $x^{2} + y^{2} - 2 x + 4 y - 20 = 0$ is

x^{2} + y^{2} - 2 x - 2 y - 2 = 0

120^{\circ}

Join BYJU'S Learning Program

Explore more

Join BYJU'S Learning Program